Differential Equations for Engineers

Lecture Notes for

Jeffrey R. Chasnov The Hong Kong University of Science and Technology Department of Mathematics Clear Water Bay, Kowloon Hong Kong

This work is licensed under the Creative Commons Attribution 3.0 Hong Kong License. To view a copy of this license, visit http://creativecommons.org/licenses/by/3.0/hk/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California, 94105, USA. Preface View the promotional video on YouTube

These are my lecture notes for my online Coursera course, Differential Equations for Engineers.I have divided these notes into chapters called Lectures, with each Lecture corresponding to a video on Coursera. I have also uploaded all my Coursera videos to YouTube, and links are placed at the top of each Lecture. There are problems at the end of each lecture chapter and I have tried to choose problems that exemplify the main idea of the lecture. Students taking a formal university course in differential equations will usually be assigned many more additional problems, but here I follow the philosophy that less is more. I give enough problems for students to solidify their understanding of the material, but not too many problems that students feel overwhelmed and drop out. I do encourage students to attempt the given problems, but if they get stuck, full solutions can be found in the Appendix. There are also additional problems at the end of coherent sections that are given as practice quizzes on the Coursera platform. Again, students should attempt these quizzes on the platform, but if a student has trouble obtaining a correct answer, full solutions are also found in the Appendix. Students who take this course are expected to know single-variable differential and integral calculus. Some knowledge of complex numbers, matrix algebra and vector calculus is required for parts of this course. Students missing this latter knowledge can ﬁnd the necessary material in the Appendix.

Jeffrey R. Chasnov Hong Kong January 2019

iii

Contents

1 Introduction to differential equations1

Practice quiz: Classify differential equations3

I First-Order Differential Equations5

2 Euler method 9

3 Separable ﬁrst-order equations 11

4 Separable ﬁrst-order equation: example 13

Practice quiz: Separable ﬁrst-order odes 15

5 Linear ﬁrst-order equations 17

6 Linear ﬁrst-order equation: example 19

Practice quiz: Linear ﬁrst-order odes 21

7 Application: compound interest 23

8 Application: terminal velocity 25

9 Application: RC circuit 27

Practice quiz: Applications 31

II Homogeneous Linear Differential Equations 33

10 Euler method for higher-order odes 37

11 The principle of superposition 39

12 The Wronskian 41

13 Homogeneous second-order ode with constant coefﬁcients 43

Practice quiz: Superposition, the Wronskian, and the characteristic equation 45

v vi CONTENTS

14 Case 1: Distinct real roots 47

15 Case 2: Complex-conjugate roots (Part A) 49

16 Case 2: Complex-conjugate roots (Part B) 51

17 Case 3: Repeated roots (Part A) 53

18 Case 3: Repeated roots (Part B) 55

Practice quiz: Homogeneous equations 57

III Inhomogeneous Linear Differential Equations 59

19 Inhomogeneous second-order ode 63

20 Inhomogeneous term: Exponential function 65

Practice quiz: Solving inhomogeneous equations 67

21 Inhomogeneous term: Sine or cosine (Part A) 69

22 Inhomogeneous term: Sine or cosine (Part B) 71

23 Inhomogeneous term: Polynomials 73

Practice quiz: Particular solutions 75

24 Resonance 77

25 Application: RLC circuit 79

26 Application: Mass on a spring 83

27 Application: Pendulum 85

28 Damped resonance 87

Practice quiz: Applications and resonance 89

IV The Laplace Transform and Series Solution Methods 91

29 Deﬁnition of the Laplace transform 95

30 Laplace transform of a constant-coefﬁcient ode 97

31 Solution of an initial value problem 99

Practice quiz: The Laplace transform method 101

32 The Heaviside step function 103

33 The Dirac delta function 105 CONTENTS vii

34 Solution of a discontinuous inhomogeneous term 107

35 Solution of an impulsive inhomogeneous term 109

Practice quiz: Discontinuous and impulsive inhomogeneous terms 111

36 The series solution method 113

37 Series solution of the Airy’s equation (Part A) 117

38 Series solution of the Airy’s equation (Part B) 121

Practice quiz: Series solutions 123

V Systems of Differential Equations 125

39 Systems of linear ﬁrst-order odes 129

40 Distinct real eigenvalues 131

41 Complex-conjugate eigenvalues 133

Practice quiz: Systems of differential equations 135

42 Phase portraits 137

43 Stable and unstable nodes 139

44 Saddle points 141

45 Spiral points 143

Practice quiz: Phase portraits 145

46 Coupled oscillators 147

47 Normal modes (eigenvalues) 149

48 Normal modes (eigenvectors) 151

Practice quiz: Normal modes 153

VI Partial Differential Equations 155

49 Fourier series 159

50 Fourier sine and cosine series 161

51 Fourier series: example 163

Practice quiz: Fourier series 165

52 The diffusion equation 167 viii CONTENTS

53 Solution of the diffusion equation (separation of variables) 169

54 Solution of the diffusion equation (eigenvalues) 171

Practice quiz: Separable partial differential equations 173

55 Solution of the diffusion equation (Fourier series) 175

56 Diffusion equation: example 177

Practice quiz: The diffusion equation 179

Appendix 180

A Complex numbers 183

B Nondimensionalization 185

C Matrices and determinants 187

D Eigenvalues and eigenvectors 189

E Partial derivatives 191

F Table of Laplace transforms 193

G Problem and practice quiz solutions 195 Lecture 1

Introduction to differential equations View this lecture on YouTube

A differential equation is an equation for a function containing derivatives of that function. For example, the differential equations for an RLC circuit, a pendulum, and a diffusing dye are given by

d2q dq 1 L + R + q = E cos ωt, (RLC circuit equation) dt2 dt C 0 d2θ dθ ml + cl + mg sin θ = F cos ωt, (pendulum equation) dt2 dt 0 ∂u ∂2u ∂2u ∂2u = D + + . (diffusion equation) ∂t ∂x2 ∂y2 ∂z2

These are second-order differential equations, categorized according to the highest order derivative. The RLC circuit equation (and pendulum equation) is an ordinary differential equation, or ode, and the diffusion equation is a partial differential equation, or pde. An ode is an equation for a function of a single variable and a pde for a function of more than one variable. A pde is theoretically equivalent to an infinite number of odes, and numerical solution of nonlinear pdes may require supercomputer resources. The RLC circuit and the diffusion equation are linear and the pendulum equation is nonlinear. In a linear differential equation, the unknown function and its derivatives appear as a linear polynomial. For instance, the general linear third-order ode, where y = y(x) and primes denote derivatives with respect to x, is given by 000 00 0 a3(x)y + a2(x)y + a1(x)y + a0(x)y = b(x), where the a and b coefficients can be any function of x. The pendulum equation is nonlinear because of the term sin θ, where θ = θ(t) is the unknown function. Making the small angle approximation, sin θ ≈ θ, the pendulum equation becomes linear. The simplest type of ode can be solved by integration. For example, a mass such as Newton’s apocryphal apple, falls downward with constant acceleration, and satisfies the differential equation

d2x = −g. dt2

With initial conditions specifying the initial height of the mass x0 and its initial velocity u0, the solution obtained by straightforward integration is given by the well-known high school physics equation

1 x(t) = x + u t − gt2. 0 0 2

1 2 LECTURE 1. INTRODUCTION TO DIFFERENTIAL EQUATIONS Practice quiz: Classify differential equations 1. By checking all that apply, classify the following differential equation: d3y d2y + y = 0 dx3 dx2 a) ﬁrst order

b) second order

c) third order

d) ordinary

e) partial

f ) linear

g) nonlinear

2. By checking all that apply, classify the following differential equation:

1 d dψ ξ2 = e−ψ ξ2 dξ dξ a) ﬁrst order

b) second order

c) ordinary

d) partial

e) linear

f ) nonlinear

3 4 LECTURE 1. INTRODUCTION TO DIFFERENTIAL EQUATIONS

3. By checking all that apply, classify the following differential equation:

∂u ∂u ∂2u + u = ν ∂t ∂x ∂x2 a) ﬁrst order

b) second order

c) ordinary

d) partial

e) linear

f ) nonlinear

4. By checking all that apply, classify the following differential equation:

d2x dx a + b + cx = 0 dt2 dt a) ﬁrst order

b) second order

c) ordinary

d) partial

e) linear

f ) nonlinear

5. By checking all that apply, classify the following differential equation:

∂2u ∂2u ∂2u ∂2u = c2 + + ∂t2 ∂x2 ∂y2 ∂z2 a) ﬁrst order

b) second order

c) ordinary

d) partial

e) linear

f ) nonlinear Solutions to the Practice quiz Week I

First-Order Differential Equations

In this week’s lectures, we discuss first-order differential equations. We begin by explaining the Euler method, which is a simple numerical method for solving an ode. Not all first-order differential equations have an analytical solution, so it is useful to understand a basic numerical method. Then the analytical solution methods for separable and linear equations are explained. We follow the dis- cussion of each theory with some simple examples. Finally, three real-world applications of first-order equations and their solutions are presented: compound interest, terminal velocity of a falling mass, and the resistor-capacitor electrical circuit. 8 Lecture 2

Euler method View this lecture on YouTube

y = y +∆x f(x ,y ) 1 0 0 0

y 1

slope = f(x ,y ) 0 0 y 0

x x 0 1

dy/dx = f (x, y) with initial condition y(x0) = y0 is integrated to x = x1 using the Euler method.

Sometimes there is no analytical solution to a ﬁrst-order differential equation and a numerical solution

must be sought. The ﬁrst-order differential equation dy/dx = f (x, y) with initial condition y(x0) = y0 provides the slope f (x0, y0) of the tangent line to the solution curve y = y(x) at the point (x0, y0). With a small step size ∆x = x1 − x0, the initial condition (x0, y0) can be marched forward to (x1, y1) along the tangent line using Euler’s method (see ﬁgure):

y1 = y0 + ∆x f (x0, y0).

This solution (x1, y1) then becomes the new initial condition and is marched forward to (x2, y2) along a newly determined tangent line with slope given by f (x1, y1). For small enough ∆x, the numerical solution converges to the unique solution, when such a solution exists. There are better numerical methods than the Euler method, but the basic principle of marching the solution forward remains the same.

9 10 LECTURE 2. EULER METHOD

Problems for Lecture 2

1. The Euler method for solving the differential equation dy/dx = f (x, y) can be rewritten in the form

k1 = ∆x f (xn, yn), yn+1 = yn + k1, and is called a ﬁrst-order Runge-Kutta method. More accurate second-order Runge-Kutta methods have the form

k1 = ∆x f (xn, yn), k2 = ∆x f (xn + α∆x, yn + βk1), yn+1 = yn + ak1 + bk2.

Some analysis (not shown here) on the second-order Runge-Kutta methods results in the constraints

a + b = 1, αb = βb = 1/2.

Write down the second-order Runge-Kutta methods corresponding to (i) a = b, and (ii) a = 0. These speciﬁc second-order Runge-Kutta methods are called the modiﬁed Euler method and the midpoint method, respectively. Solutions to the Problems Lecture 3

Separable ﬁrst-order equations View this lecture on YouTube

A ﬁrst-order ode is separable if it can be written in the form

dy g(y) = f (x), y(x ) = y , dx 0 0 where the function g(y) is independent of x and f (x) is independent of y. Integration from x0 to x results in Z x Z x g(y(x))y0(x) dx = f (x) dx. x0 x0 The integral on the left can be transformed by substituting u = y(x), du = y0(x)dx, and changing the

lower and upper limits of integration to y(x0) = y0 and y(x) = y. Therefore,

Z y Z x g(u) du = f (x) dx, y0 x0

which can often yield an analytical expression for y = y(x) if the integrals can be done and the resulting algebraic equation can be solved for y. A simpler procedure that yields the same result is to treat dy/dx as a fraction. Multiplying the differential equation by dx results directly in

g(y) dy = f (x) dx,

which is what we call a separated equation with a function of y times dy on one side, and a function of x times dx on the other side. This separated equation can then be integrated directly over y and x.

11 12 LECTURE 3. SEPARABLE FIRST-ORDER EQUATIONS

Problems for Lecture 3

1. Put the following equation in separated form. Do not integrate.

dy x2y − 4y a) = dx x + 4 dy b) = sec(y)ex−y(1 + x) dx dy xy c) = dx (x + 1)(y + 1) dθ d) + sin θ = 0 dt Solutions to the Problems Lecture 4

Separable ﬁrst-order equation: example View this lecture on YouTube

Example: Solve y0 + y2 sin x = 0, y(0) = 1.

We ﬁrst manipulate the differential equation to the form

dy = −y2 sin x dx

and then treat dy/dx as if it was a fraction to separate variables:

dy = − sin x dx. y2

We then integrate the right side from x equals 0 to x and the left side from y equals 1 to y. We obtain

Z y dy Z x = − sin x dx. 1 y2 0

Integrating, we have y x 1 − = cos x , y 1 0 or 1 1 − = cos x − 1. y Solving for y, we obtain the solution 1 y = . 2 − cos x

13 14 LECTURE 4. SEPARABLE FIRST-ORDER EQUATION: EXAMPLE

Problems for Lecture 4

1. Solve the following separable ﬁrst-order equations. √ a) dy/dx = 4x y, with y(0) = 1.

b) dx/dt = x(1 − x), with x(0) = x0 and 0 ≤ x0 ≤ 1. Solutions to the Problems Practice quiz: Separable ﬁrst-order odes √ 1. The solution of y0 = xy with initial value y(1) = 0 is given by 1 a) y(x) = (x1/2 − 1)2 9 1 b) y(x) = (x − 1)2 9 1 c) y(x) = (x3/2 − 1)2 9 1 d) y(x) = (x2 − 1)2 9 2. The solution of y2 − xy0 = 0 with initial value y(1) = 1 is given by 1 a) y(x) = 1 − ln x 1 b) y(x) = 1 − 2 ln x 1 c) y(x) = 1 + ln x 1 d) y(x) = 1 + 2 ln x 3. The solution of y0 + (sin x)y = 0 with initial value y(π/2) = 1 is given by

a) y(x) = esin x

b) y(x) = ecos x

c) y(x) = e1−sin x

d) y(x) = e1−cos x Solutions to the Practice quiz

15 16 LECTURE 4. SEPARABLE FIRST-ORDER EQUATION: EXAMPLE Lecture 5

Linear ﬁrst-order equations View this lecture on YouTube

A linear ﬁrst-order differential equation with initial condition can be written in standard form as

dy + p(x)y = g(x), y(x ) = y . (5.1) dx 0 0

All linear ﬁrst-order differential equations can be integrated using an integrating factor µ. We multiply the differential equation by the yet unknown function µ = µ(x) to obtain

dy µ(x) + p(x)y = µ(x)g(x); dx and then require µ(x) to satisfy the differential equation

dy d µ(x) + p(x)y = [µ(x)y]. (5.2) dx dx

The unknown function µ(x) is called an integrating factor because the resulting differential equation, d [µ(x)y] = µ(x)g(x), can be directly integrated. Using y(x ) = y and choosing µ(x ) = 1, we have dx 0 0 0 Z x µ(x)y − y0 = µ(x)g(x) dx; or after solving for y = y(x), x0

1 Z x y(x) = y0 + µ(x)g(x) dx . (5.3) µ(x) x0

To determine µ(x), we expand (5.2) using the product rule to obtain to yield

dy dµ dy µ + pµy = y + µ ; dx dx dx and upon canceling terms, we obtain the differential equation

dµ = p(x)µ, µ(x ) = 1. dx 0

This equation is separable and can be easily integrated to obtain

Z x µ(x) = exp p(x) dx . (5.4) x0

Equations (5.3) and (5.4) together solve the ﬁrst-order linear equation given by (5.1).

17 18 LECTURE 5. LINEAR FIRST-ORDER EQUATIONS

Problems for Lecture 5

1. Write the following linear equations in standard form. dy a) x + y = sin x; dx dy b) = x − y. dx 2. Consider the nonlinear differential equation dx/dt = x(1 − x). By deﬁning z = 1/x, show that the resulting differential equation for z is linear. Solutions to the Problems Lecture 6

Linear ﬁrst-order equation: example View this lecture on YouTube

dy Example: Solve + 2y = e−x, with y(0) = 3/4. dx

Note that this equation is not separable. With p(x) = 2 and g(x) = e−x, we have

Z x µ(x) = exp 2 dx 0 = e2x, and 3 Z x y(x) = e−2x + e2xe−x dx . 4 0 Performing the integration, we obtain

3 y(x) = e−2x + (ex − 1) , 4

which can be simpliﬁed to 1 y(x) = e−x 1 − e−x . 4

19 20 LECTURE 6. LINEAR FIRST-ORDER EQUATION: EXAMPLE

Problems for Lecture 6

1. Solve the following linear odes. dy a) = x − y, y(0) = −1; dx dy b) = 2x(1 − y), y(0) = 0. dx Solutions to the Problems Practice quiz: Linear ﬁrst-order odes 1. The solution of (1 + x2)y0 + 2xy = 2x with initial value y(0) = 0 is given by x a) y(x) = 1 + x x b) y(x) = 1 + x2 x2 c) y(x) = 1 + x x2 d) y(x) = 1 + x2 2. The solution of x2y0 = 1 − 2xy with initial value y(1) = 2 is given by 1 + x a) y(x) = x 1 + x b) y(x) = x2 1 + x2 c) y(x) = x 1 + x2 d) y(x) = x2 3. The solution of y0 + λy = a with initial value y(0) = 0 and λ > 0 is given by

a) y(x) = a(1 − eλx)

b) y(x) = a(1 − e−λx) a c) y(x) = (1 − eλx) λ a d) y(x) = (1 − e−λx) λ Solutions to the Practice quiz

21 22 LECTURE 6. LINEAR FIRST-ORDER EQUATION: EXAMPLE Lecture 7

Application: compound interest View this lecture on YouTube

The compound interest equation arises in many different engineering problems, and we consider it here as it relates to an investment. Let S(t) be the value of the investment at time t, and let r be the annual interest rate compounded after every time interval ∆t. Let k be the annual deposit amount (a negative value indicates a withdrawal), and suppose that a ﬁxed amount is deposited after every time interval ∆t. The value of the investment at the time t + ∆t is then given by

S(t + ∆t) = S(t) + (r∆t)S(t) + k∆t, (7.1) where at the end of the time interval ∆t, r∆tS(t) is the amount of interest credited and k∆t is the amount of money deposited. Rearranging the terms of (7.1) to exhibit what will soon become a derivative, we have

S(t + ∆t) − S(t) = rS(t) + k. ∆t

The equation for continuous compounding of interest and continuous deposits is obtained by taking the limit ∆t → 0. The resulting differential equation is

dS = rS + k, (7.2) dt

which can solved with the initial condition S(0) = S0, where S0 is the initial capital. The differential equation is linear and the standard form is dS/dt − rS = k, so that the integrating factor is given by

µ(t) = e−rt.

The solution is therefore

Z t rt −rt S(t) = e S0 + ke dt 0 k = S ert + ert 1 − e−rt , 0 r

where the ﬁrst term on the right-hand side comes from the initial invested capital, and the second term comes from the deposits (or withdrawals). Evidently, compounding results in the exponential growth of an investment.

23 24 LECTURE 7. APPLICATION: COMPOUND INTEREST

Problems for Lecture 7

1. Suppose a 25 year-old plans to set aside a ﬁxed amount each year until retirement at age 65. How much must he/she save each year to have $1,000,000 at retirement? Assume a 6% annual return on the saved money compounded continuously. Out of the $1,000,000, approximately how much was saved, and how much was earned on the investment?

2. A home buyer can afford to spend no more than $1500 per month on mortgage payments. Suppose that the annual interest rate is 4% and that the term of the mortgage is 30 years. Assume that interest is compounded continuously and that payments are also made continuously. Determine the maximum amount that this buyer can afford to borrow. Solutions to the Problems Lecture 8

Application: terminal velocity View this lecture on YouTube

Using Newton’s law, we model a mass m free falling under gravity but with air resistance. We assume that the force of air resistance is proportional to the speed of the mass and opposes the direction of motion. We deﬁne the x-axis to point in the upward direction, opposite the force of gravity. Near the surface of the Earth, the force of gravity is approximately constant and is given by −mg, with g = 9.8 m/s2 the usual gravitational acceleration. The force of air resistance is modeled by −kv, where v is the vertical velocity of the mass and k is a positive constant. When the mass is falling, v < 0 and the force of air resistance is positive, pointing upward and opposing the motion. The total force on the mass is therefore given by F = −mg − kv. With F = ma and a = dv/dt, we obtain the differential equation dv m = −mg − kv. (8.1) dt

The terminal velocity v∞ of the mass is deﬁned as the asymptotic velocity after air resistance balances the gravitational force. When the mass is at terminal velocity, dv/dt = 0 so that

mg v = − . (8.2) ∞ k

The approach to the terminal velocity of a mass initially at rest is obtained by solving (8.1) with initial condition v(0) = 0. The equation is linear and the standard form is dv/dt + (k/m)v = −g, so that the integrating factor is µ(t) = ekt/m, and the solution to the differential equation is

Z t v(t) = e−kt/m ekt/m(−g) dt 0 mg = − 1 − e−kt/m . k

−kt/m Therefore, v = v∞ 1 − e , and v approaches v∞ as the exponential term decays to zero.

25 26 LECTURE 8. APPLICATION: TERMINAL VELOCITY

Problems for Lecture 8

1. A male skydiver of mass m = 100 kg with his parachute closed may attain a terminal speed of 200 km/hr. How long does it take him to attain one-half his terminal speed, and how long to attain 95%? Solutions to the Problems Lecture 9

Application: RC circuit

View this lecture on YouTube

(a)

(b) + C ߝ _

RC circuit diagram.

Consider a resister R and a capacitor C connected in series as shown in the ﬁgure. A battery connects to this circuit by a switch, stepping up the voltage by E. Initially, there is no charge on the capacitor. When the switch is thrown to a, the battery connects and the capacitor charges. When the switch is thrown to b, the battery disconnects and the capacitor discharges, with energy dissipated in the resister. Here, we determine the voltage drop across the capacitor during charging and discharging. The equations for the voltage drops across a capacitor and a resister are given by

VC = q/C, VR = iR, (9.1)

where C is the capacitance and R is the resistance. The charge q and the current i are related by

dq i = . (9.2) dt

Kirchhoff’s voltage law states that the emf E in any closed loop is equal to the sum of the voltage drops in that loop. Applying Kirchhoff’s voltage law when the switch is thrown to a results in

VR + VC = E. (9.3)

Using (9.1) and (9.2), the voltage drop across the resister can be written in terms of the voltage drop across the capacitor as dV V = RC C , R dt

27 28 LECTURE 9. APPLICATION: RC CIRCUIT

and (9.3) can be rewritten to yield the linear ﬁrst-order differential equation for VC given by

dV C + V /RC = E/RC, (9.4) dt C

with initial condition VC(0) = 0. The integrating factor for this equation is

µ(t) = et/RC,

and (9.4) integrates to Z t −t/RC t/RC VC(t) = e (E/RC)e dt, 0 with solution −t/RC VC(t) = E 1 − e .

The voltage starts at zero and rises exponentially to E, with characteristic time scale given by RC. When the switch is thrown to b, application of Kirchhoff’s voltage law results in

VR + VC = 0, with corresponding differential equation

dV C + V /RC = 0. dt C

Here, we assume that the capacitance is initially fully charged so that VC(0) = E. The solution, then, during the discharge phase is given by

−t/RC VC(t) = Ee .

The voltage starts at E and decays exponentially to zero, again with characteristic time scale given by RC. 29

Problems for Lecture 9

1. Determine the current in an RC circuit during charging and discharging. Solutions to the Problems 30 LECTURE 9. APPLICATION: RC CIRCUIT Practice quiz: Applications 1. Suppose a 25 year-old plans to set aside a ﬁxed amount each year until retirement at age 65. Approximately, how much total must they have set aside to have $1,000,000 at retirement? Assume a 10% annual return on the saved money compounded continuously.

a) $75,000

b) $150,000

c) $ 225,000

d) $ 300,000

2. A male skydiver of mass 100 kg with his parachute closed may attain a terminal speed of 200 km/hr. How long does it take him to attain a speed of 150 km/hr?

a)1s

b)4s

c)8s

d) 15 s

3. An RC circuit consists of a resistor (3000 Ω) and a capacitor (0.001 F), where Ω is the ohm with units kg · m2 · s−3 · A−2, F is the Faraday with units s4 · A2 · m−2 · kg−1, and where kg is kilogram, m is meters, s is seconds, and A is amps. How long does it take for the capacitor to charge to 95% of the battery voltage?

a)1s

b)5s

c)9s

d) 14 s Solutions to the Practice quiz

31 32 LECTURE 9. APPLICATION: RC CIRCUIT Week II

Homogeneous Linear Differential Equations

This week’s lectures are on second-order differential equations. We begin by generalizing the Euler numerical method to a second-order equation, to show how numerical solutions can be obtained for equations that have no analytical solutions. We then develop two theoretical concepts used for linear equations: the principle of superposition, and the Wronskian. Armed with these concepts, we can then ﬁnd analytical solutions to a homogeneous second-order ode with constant coefﬁcients. We make use of an exponential ansatz, and convert the differential equation to a quadratic equation called the characteristic equation of the ode. The characteristic equation may have real or complex roots and we discuss the solutions for these different cases. 36 Lecture 10

Euler method for higher-order odes View this lecture on YouTube

In practice, most higher-order odes are solved numerically. For example, consider the general second- order ode given by x¨ = f (t, x, x˙).

Here, we adopt the widely used physics notation x˙ = dx/dt and x¨ = d2x/dt2. The dot notation is used to represent only time derivatives and can be applied to any dependent variable that is a function of time. To solve numerically, we convert the second-order ode to a pair of first-order odes. Define u = x˙, and write the first-order system as

x˙ = u, (10.1) u˙ = f (t, x, u). (10.2)

The ﬁrst ode, (10.1), gives the slope of the tangent line to the curve x = x(t). The second ode, (10.2), gives the slope of the tangent line to the curve u = u(t). Beginning at the initial values (x, u) = (x0, u0) at the time t = t0, we move along the tangent lines to determine x1 = x(t0 + ∆t) and u1 = u(t0 + ∆t):

x1 = x0 + ∆tu0,

u1 = u0 + ∆t f (t0, x0, u0).

The values x1 and u1 at the time t1 = t0 + ∆t are then used as new initial values to march the solution forward to time t2 = t1 + ∆t. When a unique solution of the ode exists, the numerical solution converges to this unique solution as ∆t → 0.

37 38 LECTURE 10. EULER METHOD FOR HIGHER-ORDER ODES

Problems for Lecture 10

1. Write the second-order ode, θ¨ + θ˙/q + sin θ = f cos ωt, as a system of two ﬁrst-order odes.

2. Write down the second-order Runge-Kutta modiﬁed Euler method (predictor-corrector method) for the following system of two ﬁrst-order odes:

x˙ = f (t, x, y), y˙ = g(t, x, y).

Solutions to the Problems Lecture 11

The principle of superposition View this lecture on YouTube

Consider the homogeneous linear second-order ode given by

x¨ + p(t)x˙ + q(t)x = 0; and suppose that x = X1(t) and x = X2(t) are solutions. We consider a linear combination of X1 and X2 by letting x = c1X1(t) + c2X2(t),

with c1 and c2 constants. The principle of superposition states that x is also a solution to the homogeneous ode. To prove this, we compute

x¨ + px˙ + qx = c1X¨ 1 + c2X¨ 2 + p c1X˙ 1 + c2X˙ 2 + q (c1X1 + c2X2) = c1 X¨ 1 + pX˙ 1 + qX1 + c2 X¨ 2 + pX˙ 2 + qX2

= c1 × 0 + c2 × 0 = 0,

since X1 and X2 were assumed to be solutions of the homogeneous ode. We have therefore shown that any linear combination of solutions to the homogeneous linear second-order ode is also a solution.

39 40 LECTURE 11. THE PRINCIPLE OF SUPERPOSITION

Problems for Lecture 11

1. Consider the inhomogeneous linear second-order ode given by

x¨ + p(t)x˙ + q(t)x = g(t); and suppose that x = xh(t) is a solution of the homogeneous equation and x = xp(t) is a solution of the inhomogeneous equation. Prove that x = xh(t) + xp(t) is a solution of the inhomogeneous equation. Solutions to the Problems Lecture 12

The Wronskian View this lecture on YouTube

Suppose that we have determined that x = X1(t) and x = X2(t) are solutions to

x¨ + p(t)x˙ + q(t)x = 0, and that we now attempt to write the general solution to the ode as

x = c1X1(t) + c2X2(t).

We must then ask whether this general solution can satisfy the two initial conditions given by

x(t0) = x0, x˙(t0) = u0.

Applying these initial conditions to our proposed general solution, we obtain

c1X1(t0) + c2X2(t0) = x0, c1X˙ 1(t0) + c2X˙ 2(t0) = u0, which is a system of two linear equations for c1 and c2. A unique solution exists provided

( ) ( ) X1 t0 X2 t0 ˙ ˙ W = = X1(t0)X2(t0) − X1(t0)X2(t0) 6= 0, X˙ 1(t0) X˙ 2(t0) where the determinant given by W is called the Wronskian.

For example, with ω 6= 0, the two solutions X1(t) = cos ωt and X2(t) = sin ωt have a nonzero Wronskian for all t since

W = (cos ωt)(ω cos ωt) − (−ω sin ωt)(sin ωt) = ω.

When the Wronskian is not equal to zero, we say that the two solutions X1(t) and X2(t) are linearly independent. The concept of linear independence is borrowed from linear algebra, and indeed, the set of all functions that satisfy a second-order linear homogeneous ode can be shown to form a two- dimensional vector space.

41 42 LECTURE 12. THE WRONSKIAN

Problems for Lecture 12

1. Show that X1(t) = exp (αt) and X2(t) = exp (βt) have a nonzero Wronskian for all t provided α 6= β. Solutions to the Problems Lecture 13

Homogeneous second-order ode with constant coefﬁcients View this lecture on YouTube

We now study solutions of the homogeneous constant-coefﬁcient ode, written as

ax¨ + bx˙ + cx = 0, with a, b, and c constants. Our solution method ﬁnds two linearly independent solutions, multiplies each of these solutions by a constant, and adds them. The two free constants are then determined from initial values on x and x˙. Because of the differential properties of the exponential function, a natural ansatz, or educated guess, for the form of the solution is x = ert, where r is a constant to be determined. Successive differentiation results in x˙ = rert and x¨ = r2ert, and substitution into the ode yields

ar2ert + brert + cert = 0.

Our choice of exponential function is now rewarded by the explicit cancelation of ert. The result is a quadratic equation for the unknown constant r:

ar2 + br + c = 0.

Our ansatz has thus converted a differential equation for x into an algebraic equation for r. This algebraic equation is called the characteristic equation of the ode. Using the quadratic formula, the two solutions of the characteristic equation are given by √ −b ± b2 − 4ac r± = . 2a

There are three cases to consider: (1) if b2 − 4ac > 0, the roots are distinct and real ; (2) if b2 − 4ac < 0, the roots are distinct and complex-conjugates; and (3) if b2 − 4ac = 0, the roots are repeated. We will consider these three cases in turn.

43 44 LECTURE 13. HOMOGENEOUS SECOND-ORDER ODE WITH CONSTANT COEFFICIENTS

Problems for Lecture 13

1. For the following differential equations, determine the roots of the characteristic equation.

a) x¨ − x = 0;

b) x¨ + x = 0;

c) x¨ − 2x˙ + x = 0. Solutions to the Problems Practice quiz: Superposition, the Wronskian, and the characteristic equation 1. Which two functions have a nonzero Wronskian?

a)2 e2t, 3e2t

b) et, et−t0

c) sin t, sin (t − π/2)

d) sin t, sin (t − π)

2. Let x = X1(t) and x = X2(t) be solutions to a homogeneous linear second-order ode. Which of the following functions is generally not a solution? 1 a) x = (X + X ) 2 1 2 1 b) x = (X − X ) 2 1 2 c) x = 0

d) x = X1(t)X2(t)

3. Which of the following odes has a characteristic equation with complex-conjugate roots? I. x¨ + x˙ + x = 0 II. x¨ + x˙ − x = 0 III. x¨ − x˙ + x = 0 IV. x¨ − x˙ − x = 0

a) I only

b) IV only

c) I and III

d) II and IV Solutions to the Practice quiz

45 46 LECTURE 13. HOMOGENEOUS SECOND-ORDER ODE WITH CONSTANT COEFFICIENTS Lecture 14

Case 1: Distinct real roots View this lecture on YouTube

When the roots of the characteristic equation are distinct and real, then the general solution to the second-order ode can be written as a linear superposition of the two solutions er1t and er2t; that is,

r1t r2t x(t) = c1e + c2e .

The unknown constants c1 and c2 can then be determined by the given initial conditions x(t0) = x0 and x˙(t0) = u0.

Example: Solve x¨ + 5x˙ + 6x = 0 with x(0) = 2, x˙(0) = 3.

We take as our ansatz x = ert and obtain the characteristic equation

r2 + 5r + 6 = 0,

which factors to (r + 3)(r + 2) = 0.

The general solution to the ode is thus

−2t −3t x(t) = c1e + c2e .

The solution for x˙ obtained by differentiation is

−2t −3t x˙(t) = −2c1e − 3c2e .

Use of the initial conditions then results in two equations for the two unknown constant c1 and c2,

c1 + c2 = 2, −2c1 − 3c2 = 3, with solution c1 = 9 and c2 = −7. Therefore, the unique solution that satisﬁes both the ode and the initial conditions is

x(t) = 9e−2t − 7e−3t 7 = 9e−2t 1 − e−t . 9

47 48 LECTURE 14. CASE 1: DISTINCT REAL ROOTS

Problems for Lecture 14

1. Solve the following initial value problem: x¨ + 4x˙ + 3x = 0, with x(0) = 1 and x˙(0) = 0.

2. Find the solution of x¨ − x = 0, with x(0) = x0 and x˙(0) = u0. Solutions to the Problems Lecture 15

Case 2: Complex-conjugate roots (Part A) View this lecture on YouTube

When the roots of the characteristic equation are complex conjugates, we can deﬁne real numbers λ and µ such that the two roots are given by

r = λ + iµ, r¯ = λ − iµ.

We have thus found the following two complex exponential solutions to the differential equation:

z(t) = eλteiµt, z¯(t) = eλte−iµt.

Applying the principle of superposition, any linear combination of z and z¯ is also a solution to the second-order ode, and we can form two different linear combinations of z(t) and z¯(t) that are real, namely x1(t) = Re z(t) and x2(t) = Im z(t). We have

λt λt x1(t) = e cos µt, x2(t) = e sin µt.

Having found these two real solutions, we can then apply the principle of superposition a second time to determine the general solution for x(t):

x(t) = eλt (A cos µt + B sin µt) .

The real part of the roots of the characteristic equation appear in the exponential function, the imaginary part appears in the cosine and sine.

49 50 LECTURE 15. CASE 2: COMPLEX-CONJUGATE ROOTS (PART A) Lecture 16

Case 2: Complex-conjugate roots (Part B) View this lecture on YouTube

Example: Solve x¨ + x˙ + x = 0 with x(0) = 1 and x˙(0) = 0.

The characteristic equation is r2 + r + 1 = 0, with roots √ √ 1 3 1 3 r = − + i , r¯ = − − i . 2 2 2 2

The general solution of the ode is therefore √ √ ! 3 3 x(t) = e−t/2 A cos t + B sin t . 2 2

The derivative is √ √ ! √ √ √ ! 1 3 3 3 3 3 x˙(t) = − e−t/2 A cos t + B sin t + e−t/2 −A sin t + B cos t . 2 2 2 2 2 2

Applying the initial conditions x(0) = 1 and x˙(0) = 0 results in √ 1 3 A = 1, − A + B = 0, 2 2 √ with solution A = 1 and B = 3/3. Therefore, the resulting solution is √ √ √ ! 3 3 3 x(t) = e−t/2 cos t + sin t . 2 3 2

51 52 LECTURE 16. CASE 2: COMPLEX-CONJUGATE ROOTS (PART B)

Problems for Lecture 16

1. Solve x¨ − 2x˙ + 5x = 0, with x(0) = 1 and x˙(0) = 0.

2. Solve x¨ + x = 0, with x(0) = x0 and x˙(0) = u0. Solutions to the Problems Lecture 17

Case 3: Repeated roots (Part A) View this lecture on YouTube

If the exponential ansatz yields only one linearly independent solution, we need to ﬁnd the missing second solution. We will do this by starting with the roots of the characteristic equation given by r = λ ± iµ, and taking the limit as µ goes to zero. Now, the general solution for the case of complex-conjugate roots is given by

x(t) = eλt (A cos µt + B sin µt) , and to limit this solution as µ → 0 requires ﬁrst satisfying the general initial conditions x(0) = x0 and x˙(0) = u0. Solving for A and B, the solution becomes

u − λx x(t; µ) = eλt x cos µt + 0 0 sin µt , 0 µ where we have written x = x(t; µ) to show explicitly that x now depends on the parameter µ. −1 Taking the limit as µ → 0, and using limµ→0 µ sin µt = t, we have

λt lim x(t; µ) = e x0 + (u0 − λx0)t ; µ→0 and the second solution is observed to be a constant, u0 − λx0, times t times the ﬁrst solution. The general solution for the case of repeated roots can therefore be written in the form

rt x(t) = (c1 + c2t)e , where r is the repeated root. The main result is that the second solution is t times the ﬁrst solution.

53 54 LECTURE 17. CASE 3: REPEATED ROOTS (PART A) Lecture 18

Case 3: Repeated roots (Part B) View this lecture on YouTube

Example: Solve x¨ + 2x˙ + x = 0 with x(0) = 1 and x˙(0) = 0.

The characteristic equation is r2 + 2r + 1 = (r + 1)2 = 0, which has a repeated root given by r = −1. Therefore, the general solution to the ode is

−t −t x(t) = (c1 + c2t)e , x˙(t) = (c2 − c1 − c2t)e .

Applying the initial conditions, we have c1 = 1 and c2 − c1 = 0, so that c1 = c2 = 1. Therefore, the solution is x(t) = (1 + t)e−t.

55 56 LECTURE 18. CASE 3: REPEATED ROOTS (PART B)

Problems for Lecture 18

1. Solve x¨ − 2x˙ + x = 0, with x(0) = 1 and x˙(0) = 0. Solutions to the Problems Practice quiz: Homogeneous equations 1. The solution of x¨ − 3x˙ + 2x = 0 with initial values x(0) = 1 and x˙(0) = 0 is given by

a) e2t(2 − e−t)

b) −e2t(1 − 2e−t) 1 c) e3t(3 − e−2t) 2 1 d) − e3t(1 − 3e−2t) 2 2. The solution of x¨ − 2x˙ + 2x = 0 with initial values x(0) = 1 and x˙(0) = 0 is given by

a) et(cos t + sin t)

b) et(cos t − sin t) 1 c) et(cos 2t − sin 2t) 2 d) e2t(cos t − 2 sin 2t)

3. The solution of x¨ + 2x˙ + x = 0 with initial values x(0) = 0 and x˙(0) = 1 is given by

a) te−t

b) e−t(1 − t)

c) tet

d) et(1 − t) Solutions to the Practice quiz

57 58 LECTURE 18. CASE 3: REPEATED ROOTS (PART B) Week III

Inhomogeneous Linear Differential Equations

We now add an inhomogeneous term to the second-order ode with constant coefﬁcients. The inhomogeneous term may be an exponential, a sine or cosine, or a polynomial. We further study the phenomena of resonance, when the forcing frequency is equal to the natural frequency of the oscillator. Finally, we learn about three important applications: the RLC electrical circuit, a mass on a spring, and the pendulum. 62 Lecture 19

Inhomogeneous second-order ode View this lecture on YouTube

The inhomogeneous linear second-order ode is given by

x¨ + p(t)x˙ + q(t)x = g(t),

with initial conditions x(t0) = x0 and x˙(t0) = u0. There is a three-step solution method when the inhomogeneous term g(t) 6= 0. (1) Solve the homogeneous ode

x¨ + p(t)x˙ + q(t)x = 0, for two independent solutions x = x1(t) and x = x2(t), and form a linear superposition to obtain the general solution

xh(t) = c1x1(t) + c2x2(t), where c1 and c2 are free constants. (2) Find a particular solution x = xp(t) that solves the inhomogeneous ode. A particular solution is most easily found when p(t) and q(t) are constants, and when g(t) is a combination of polynomials, exponentials, sines and cosines. (3) Write the general solution of the inhomogeneous ode as the sum of the homogeneous and particular solutions,

x(t) = xh(t) + xp(t), and apply the initial conditions to determine the constants c1 and c2 appearing in the homogeneous solution. Note that because of the linearity of the differential equations,

d2 d x¨ + px˙ + qx = (x + x ) + p (x + x ) + q(x + x ) dt2 h p dt h p h p

= (x¨h + px˙h + qxh) + (x¨p + px˙ p + qxp) = 0 + g = g, so that the sum of the homogeneous and particular solutions solve the ode, and the two free constants in xh can be used to satisfy the two initial conditions.

63 64 LECTURE 19. INHOMOGENEOUS SECOND-ORDER ODE

Problems for Lecture 19

1. Consider the inhomogeneous linear second-order ode given by

x¨ + p(t)x˙ + q(t)x = g1(t) + g2(t).

Show that

x(t) = xh(t) + xp1 (t) + xp2 (t)

is the general solution, where xh(t) is the general solution to the homogeneous ode, xp1 (t) is a par-

ticular solution for the inhomogeneous ode with only g1(t) on the right-hand-side, and xp2 (t) is a particular solution for the inhomogeneous ode with only g2(t) on the right-hand side. Solutions to the Problems Lecture 20

Inhomogeneous term: Exponential function View this lecture on YouTube Example: Solve x¨ − 3x˙ − 4x = 3e2t with x(0) = 1 and x˙(0) = 0. First, we solve the homogeneous equation. The characteristic equation is

r2 − 3r − 4 = (r − 4)(r + 1) = 0,

so that 4t −t xh(t) = c1e + c2e .

Second, we ﬁnd a particular solution of the inhomogeneous equation. We try an ansatz such that the exponential function cancels: x(t) = Ae2t, where A is an undetermined coefﬁcient. Substituting our ansatz into the ode, we obtain

4A − 6A − 4A = 3, from which we determine A = −1/2. Obtaining a solution for A independent of t justiﬁes the ansatz. Third, we write the general solution to the ode as the sum of the homogeneous and particular solutions, and determine c1 and c2 that satisfy the initial conditions. We have

1 x(t) = c e4t + c e−t − e2t, x˙(t) = 4c e4t − c e−t − e2t. 1 2 2 1 2

Applying the initial conditions yields the two linear equations

3 c + c = , 4c − c = 1, 1 2 2 1 2

with solution c1 = 1/2 and c2 = 1. Therefore, the solution for x(t) that satisﬁes both the ode and the initial conditions is given by

1 1 1 x(t) = e4t − e2t + e−t = e4t 1 − e−2t + 2e−5t , 2 2 2 where we have grouped the terms in the ﬁnal solution to better display the asymptotic behavior for large t.

65 66 LECTURE 20. INHOMOGENEOUS TERM: EXPONENTIAL FUNCTION

Problems for Lecture 20

1. Solve x¨ + 5x˙ + 6x = e−t, with x(0) = 0 and x˙(0) = 0. Solutions to the Problems Practice quiz: Solving inhomogeneous equations 1. All the questions will consider the differential equation given by

x¨ + 5x˙ + 6x = 2e−t.

What is the solution that satisﬁes x(0) = 0 and x˙(0) = 0?

a) x(t) = e−t(1 − 2e−t + e−2t)

b) x(t) = et(1 − 4e−3t + 3e−4t)

c) x(t) = e3t(3 − 4e−t + e−4t)

d) x(t) = e3t(1 − 2e−t + e−2t)

2. All the questions will consider the differential equation given by

x¨ + 5x˙ + 6x = 2e−t.

What is the solution that satisﬁes x(0) = 1 and x˙(0) = 0?

a) x(t) = e−t(1 + e−t − e−2t)

b) x(t) = et(1 − e−3t + e−4t)

c) x(t) = e3t(1 − e−t + e−4t)

d) x(t) = −e3t(1 − e−t − e−2t)

3. All the questions will consider the differential equation given by

x¨ + 5x˙ + 6x = 2e−t.

What is the solution that satisﬁes x(0) = 0 and x˙(0) = 1?

a) x(t) = e−t(1 − e−t)

b) x(t) = et(1 − 3e−3t + 2e−4t)

c) x(t) = e3t(4 − 5e−t + e−4t)

d) x(t) = e3t(2 − 3e−t + e−2t) Solutions to the Practice quiz

67 68 LECTURE 20. INHOMOGENEOUS TERM: EXPONENTIAL FUNCTION Lecture 21

Inhomogeneous term: Sine or cosine (Part A) View this lecture on YouTube

Example: Find a particular solution of x¨ − 3x˙ − 4x = 2 sin t. We will demonstrate two solution methods in this and the next lecture. The ﬁrst method tries the ansatz x(t) = A cos t + B sin t, where the argument of cosine and sine must agree with the argument of sine in the inhomogeneous term. The cosine term is required because the derivative of sine is cosine. Upon substitution into the differential equation, we obtain

(−A cos t − B sin t) − 3 (−A sin t + B cos t) − 4 (A cos t + B sin t) = 2 sin t,

or regrouping terms, − (5A + 3B) cos t + (3A − 5B) sin t = 2 sin t.

This equation is valid for all t, and in particular for t = 0 and π/2, for which the sine and cosine functions vanish. For these two values of t, we ﬁnd

5A + 3B = 0, 3A − 5B = 2;

and solving we obtain A = 3/17 and B = −5/17. The particular solution is therefore given by

1 x = (3 cos t − 5 sin t) . p 17

69 70 LECTURE 21. INHOMOGENEOUS TERM: SINE OR COSINE (PART A) Lecture 22

Inhomogeneous term: Sine or cosine (Part B) View this lecture on YouTube Example: Find a particular solution of x¨ − 3x˙ − 4x = 2 sin t. The second method makes use of the relation eit = cos t + i sin t to convert the sine inhomogeneous term to an exponential function. We introduce the complex function z = z(t) and note that sin t = Im{eit}. We then consider the complex ode

z¨ − 3z˙ − 4z = 2eit, where x = Im{z} satisﬁes the original differential equation for x. To ﬁnd a particular solution of the complex equation, we try the ansatz z(t) = Ceit, where we now expect C to be a complex constant. Upon substitution into the complex ode, and using i2 = −1, we obtain −C − 3iC − 4C = 2; or upon solving for C,

−2 −5 + 3i C = = . 5 + 3i 17

Therefore, 1 1 x = Im{z } = Im (−5 + 3i)(cos t + i sin t) = (3 cos t − 5 sin t), p p 17 17 the same result as in the ﬁrst method.

71 72 LECTURE 22. INHOMOGENEOUS TERM: SINE OR COSINE (PART B)

Problems for Lecture 22

1. Find a particular solution of x¨ − 3x˙ − 4x = 2 cos t by trying a

a) cosine and sine ansatz;

b) an exponential ansatz. Solutions to the Problems Lecture 23

Inhomogeneous term: Polynomials View this lecture on YouTube Example: Find a particular solution of x¨ + x˙ − 2x = t2. Our ansatz should be a polynomial in t of the same order as the inhomogeneous term. Accordingly, we try x(t) = At2 + Bt + C.

Upon substitution into the ode, we have

2A + (2At + B) − 2(At2 + Bt + C) = t2,

or −2At2 + 2(A − B)t + (2A + B − 2C)t0 = t2.

Equating powers of t, −2A = 1, 2(A − B) = 0, 2A + B − 2C = 0;

and solving, 1 1 3 A = − , B = − , C = − . 2 2 4 The particular solution is therefore

1 1 3 x (t) = − t2 − t − . p 2 2 4

73 74 LECTURE 23. INHOMOGENEOUS TERM: POLYNOMIALS

Problems for Lecture 23

1. Find a particular solution of x¨ + x˙ + x = t. Solutions to the Problems Practice quiz: Particular solutions 1. A particular solution of x¨ + 3x˙ + 2x = 2e2t is given by 1 a) x (t) = e−2t p 6 1 b) x (t) = e2t p 6 1 c) x (t) = − e−2t p 6 1 d) x (t) = − e2t p 6 2. A particular solution of x¨ − x˙ − 2x = 2 cos 2t is given by 1 a) x (t) = (cos 2t + 3 sin 2t) p 10 1 b) x (t) = − (cos 2t + 3 sin 2t) p 10 1 c) x (t) = (3 cos 2t + sin 2t) p 10 1 d) x (t) = − (3 cos 2t + sin 2t) p 10 3. A particular solution of x¨ − 3x˙ + 2x = t + 1 is given by 5 1 a) x (t) = t + p 4 2 5 1 b) x (t) = t − p 4 2 1 5 c) x (t) = t + p 2 4 1 5 d) x (t) = t − p 2 4 Solutions to the Practice quiz

75 76 LECTURE 23. INHOMOGENEOUS TERM: POLYNOMIALS Lecture 24

Resonance View this lecture on YouTube

Resonance occurs when the forcing frequency ω matches the natural frequency ω0 of an oscillator. To illustrate resonance, we consider the inhomogeneous linear second-order ode

2 x¨ + ω0 x = f cos ωt, x(0) = 0, x˙(0) = 0; and determine what happens to the solution in the limit ω → ω0. 2 2 The homogeneous equation has characteristic equation r + ω0 = 0, with solution r = ±iω0, so that the solution to the homogeneous equation is

xh(t) = c1 cos ω0t + c2 sin ω0t.

A particular solution of the equation can be found by trying x(t) = A cos ωt. Upon substitution into 2 2 the ode, we can solve for A to obtain A = f /(ω0 − ω ), so that

f x (t) = cos ωt. p 2 2 ω0 − ω

Our general solution is thus

f x(t) = c cos ω t + c sin ω t + cos ωt. 1 0 2 0 2 2 ω0 − ω

2 2 Initial conditions are satisﬁed when c1 = − f /(ω0 − ω ) and c2 = 0, so that

f (cos ωt − cos ω t) x(t) = 0 . 2 2 ω0 − ω

Resonance occurs in the limit ω → ω0; that is, the frequency of the inhomogeneous term (the external force) matches the frequency of the homogeneous solution (the free oscillation). By L’Hospital’s rule, the limit of x = x(t) is found by differentiating with respect to ω:

f (cos ωt − cos ω t) − f t sin ωt f t sin ω t lim x(t) = lim 0 = lim = 0 . ω→ω ω→ω 2 2 ω→ω − 0 0 ω0 − ω 0 2ω 2ω0

At resonance, the oscillation increases linearly with t, resulting in a large amplitude oscillation. More generally, if the inhomogeneous term in the differential equation is a solution of the corresponding homogeneous differential equation, then one should multiply the usual ansatz for the particular solution by t.

77 78 LECTURE 24. RESONANCE

Problems for Lecture 24

1. Solve x¨ + 3x˙ + 2x = e−2t, with x(0) = 0 and x˙(0) = 0. Observe that the inhomogeneous term is a solution of the homogeneous equation. To ﬁnd a particular solution, the usual exponential ansatz must be multiplied by t. Solutions to the Problems Lecture 25

Application: RLC circuit View this lecture on YouTube

C ~ 휀(푡)

RLC circuit diagram

Consider a resister R, an inductor L and a capacitor C connected in series as shown in the above ﬁgure.

An AC generator provides a time-varying electromotive force to the circuit, given by E(t) = E0 cos ωt. The equations for the voltage drops across a capacitor, a resister and an inductor are

di V = q/C, V = iR, V = L, C R L dt

where C is the capacitance, R is the resistance and L is the inductance. The charge q and the current i are related by i = dq/dt. Kirchhoff’s voltage law states that the electromotive force applied to any closed loop is equal to the

sum of the voltage drops in that loop. Appying Kirchhoff’s law, we have VL + VR + VC = E(t), or

d2q dq 1 L + R + q = E cos ωt. (25.1) dt2 dt C 0

This is a second-order linear inhomogeneous differential equation with constant coefficients. To reduce the number of free parameters in this equation, we can nondimensionalize. We first define the natural frequency of oscillation of a system to be the frequency of oscillation in the absence of any driving or damping forces. For the RLC circuit, the natural frequency of oscillation is given √ by ω0 = 1/ LC, and making use of ω0, we can define a dimensionless time τ and a dimensionless charge Q by 2 ω0 L τ = ω0t, Q = q. E0

79 80 LECTURE 25. APPLICATION: RLC CIRCUIT

The resulting dimensionless equation for the RLC circuit can then be found to be

d2Q dQ + α + Q = cos βτ, (25.2) dτ2 dτ

where α and β are dimensionless parameters given by

R ω α = , β = . Lω0 ω0

Notice that the original ﬁve parameters in (25.1), namely R, L, C, E0 and ω, have been reduced to the two dimensionless parameters α and β in (25.2). 81

Problems for Lecture 25

1. Determine how to nondimensionalize the RLC circuit equation so that the dimensionless equation takes the form d2Q dQ α + + Q = cos βτ. dτ2 dτ What are the deﬁnitions of τ, Q, α and β? Solutions to the Problems 82 LECTURE 25. APPLICATION: RLC CIRCUIT Lecture 26

Application: Mass on a spring View this lecture on YouTube

Mass-spring system

Consider a mass connected by a spring to a ﬁxed wall, as shown above. The spring force is modeled

by Hooke’s law, Fs = −kx, and sliding friction is modeled as Ff = −cdx/dt. An external force is applied and is assumed to be sinusoidal, with Fe = F0 cos ωt. Newton’s equation, F = ma, results in

d2x dx m + c + kx = F cos ωt, dt2 dt 0

another second-order linear inhomogeneous differential equation with constant coefﬁcients. Here, the √ natural frequency of oscillation is given by ω0 = k/m, and we deﬁne a dimensionless time τ and a dimensionless position X by 2 mω0 τ = ω0t, X = x. F0 The resulting dimensionless equation is given by

d2X dX + α + X = cos βτ, (26.1) dτ2 dτ where here, α and β are dimensionless parameters given by

c ω α = , β = . mω0 ω0

Even though the physical problem is different, the resulting dimensionless equation for the mass- spring system is the same as that for the RLC circuit.

83 84 LECTURE 26. APPLICATION: MASS ON A SPRING

Problems for Lecture 26

1. Determine how to nondimensionalize the mass on a spring equation so that the dimensionless equation takes the form d2X dX α + + X = cos βτ. dτ2 dτ What are the deﬁnitions of τ, X, α and β? Solutions to the Problems Lecture 27

Application: Pendulum View this lecture on YouTube

θ l

mg sin θ θ mg

The pendulum

Consider a mass attached to a massless rigid rod and constrained to move along an arc of a circle centered at the pivot point (see ﬁgure above). Suppose l is the ﬁxed length of the connecting rod, and θ is the angle it makes with the vertical. We apply Newton’s equation, F = ma, to the mass with origin at the bottom and axis along the arc with positive direction to the right. The position s of the mass along the arc is given by s = lθ. The relevant gravitational force on the pendulum is the component along the arc, and is given by

Fg = −mg sin θ. We model friction to be proportional to the velocity of the pendulum along the arc, that is Ff = −cs˙ = −clθ˙. With a sinusoidal external force, Fe = F0 cos ωt, Newton’s equation ms¨ = Fg + Ff + Fe results in mlθ¨ + clθ˙ + mg sin θ = F0 cos ωt.

At small amplitudes of oscillation, we can approximate sin θ ≈ θ, and the natural frequency of p oscillation ω0 of the mass is given by ω0 = g/l. Nondimensionalizing time as τ = ω0t, the dimensionless pendulum equation becomes

d2θ dθ + α + sin θ = γ cos βτ, dτ2 dτ

where α, β, and γ are dimensionless parameters given by

c ω F = = = 0 α , β , γ 2 . mω0 ω0 mlω0

85 86 LECTURE 27. APPLICATION: PENDULUM

The nonlinearity of the pendulum equation, with the term sin θ, results in the additional dimensionless parameter γ. For small amplitude of oscillation, however, we can scale θ by θ = γΘ, and the small amplitude dimensionless equation becomes

d2Θ dΘ + α + Θ = cos βτ, (27.1) dτ2 dτ the same equation as the dimensionless equations for the RLC circuit and the mass on a spring. Lecture 28

Damped resonance View this lecture on YouTube The dimensionless odes for the RLC circuit, mass on a spring, and small amplitude pendulum have the form x¨ + αx˙ + x = cos βt, (28.1)

> where the physical constraints of these applications require that α 0. The corresponding√ homoge- 2 neous equation has the characteristic polynomial r + αr + 1 = 0, with roots r± = (−α ± α2 − 4)/2. No matter the sign of the discriminant, we have Re{r±} < 0. Therefore, both linearly independent homogeneous solutions decay exponentially to zero, and the long-time solution of the differential equation reduces to the non-decaying particular solution. Since the initial conditions are satisﬁed by the free constants multiplying the decaying homogeneous solutions, the long-time solution is independent of the initial conditions. If we are only interested in the long-time solution of (28.1), we only need determine the particular solution. We can consider the complex ode

z¨ + αz˙ + z = eiβt,

iβt 2 with xp = Re(zp). With the ansatz zp = Ae , we have −β A + iαβA + A = 1, or

1 A = . (1 − β2) + iαβ

Of particular interest is when the forcing frequency ω is equal to the natural frequency ω0 of the undampled oscillator. In the dimensionless equation of (28.1), this resonance situation corresponds to β = 1. In this case, A = 1/iα and

it xp = Re{e /iα} = (1/α) sin t.

The oscillator position is observed to be π/2 out of phase with the external force, or in other words, the velocity of the oscillator, not the position, is in phase with the force. Also, the amplitude of oscillation is seen to be inversely proportional to the damping coefﬁcient represented by α. Smaller and smaller damping will obviously lead to larger and larger oscillations.

87 88 LECTURE 28. DAMPED RESONANCE

Problems for Lecture 28

1. Consider the differential equation

x¨ + αx˙ + x = cos βt.

Find the long-time amplitude of oscillation as a function of α and β. Solutions to the Problems Practice quiz: Applications and resonance 1. The solution of the differential equation x¨ + x˙ = 1, with x(0) = 0 and x˙(0) = 0, is given by

a) t(1 − et)

b) t(1 − e−t)

c) (t + 1) − et

d) (t − 1) + e−t

2. The solution of the differential equation x¨ − x = cosh t, with x(0) = 0 and x˙(0) = 0, is given by 1 a) t cosh t 2 1 b) t sinh t 2 1 c) (t cosh t + sinh t) 2 1 d) (t cosh t − sinh t) 2 3. When comparing the RLC circuit to the mass on a spring and to the pendulum, the resistor R plays the role of

a) mass

b) gravity

c) friction

d) restoring force Solutions to the Practice quiz

89 90 LECTURE 28. DAMPED RESONANCE Week IV

The Laplace Transform and Series Solution Methods

We present here two new analytical solution methods for solving linear differential equations. The ﬁrst is the Laplace transform method, which is used to solve the constant-coefﬁcient ode with a discontinuous or impulsive inhomogeneous term. The Laplace transform is a good vehicle in general for introducing sophisticated integral transform techniques within an easily understandable context. We also discuss the series solution of a linear differential equation. Although we do not go deeply here, an introduction to this technique may be useful to students that encounter it again in more advanced courses. 94 Lecture 29

Deﬁnition of the Laplace transform View this lecture on YouTube

The Laplace transform of f (t), denoted by F(s) = L{ f (t)}, is deﬁned by the integral

Z ∞ F(s) = e−st f (t) dt. 0

The values of s may be restricted to ensure convergence. The Laplace transform can be shown to be a linear transformation. We have

Z ∞ −st L{c1 f1(t) + c2 f2(t)} = e c1 f1(t) + c2 f2(t) dt 0 Z ∞ Z ∞ −st −st = c1 e f1(t) dt + c2 e f2(t) dt 0 0

= c1L{ f1(t)} + c2L{ f2(t)}.

There is also a one-to-one correspondence between functions and their Laplace transforms, and a table of Laplace transforms is used to ﬁnd both Laplace and inverse Laplace transforms of commonly occurring functions. To construct such a table, integrals have been performed, where for example, we can compute

Z ∞ L{eat} = e−(s−a)t dt 0 1 ∞ = − e−(s−a)t s − a 0 1 = . s − a

Here, the assumption is made that s > a. A Table of Laplace transforms can be found in AppendixF.

95 96 LECTURE 29. DEFINITION OF THE LAPLACE TRANSFORM

Problems for Lecture 29

1. Compute the Laplace transform of f (t) = sin bt. Solutions to the Problems Lecture 30

Laplace transform of a constant-coefﬁcient ode View this lecture on YouTube

Consider the inhomogeneous constant-coefﬁcient second-order ode, given by

ax¨ + bx˙ + cx = g(t), x(0) = x0, x˙(0) = u0.

We Laplace transform the ode making use of the linearity of the transform:

aL{x¨} + bL{x˙} + cL{x} = L{g}.

We deﬁne X(s) = L{x(t)} and G(s) = L{g(t)}. The Laplace transforms of the derivatives can be found from integrating by parts. We have

Z ∞ ∞ Z ∞ −st −st −st e x˙ dt = xe + s e x dt = sX(s) − x0; 0 0 0

and Z ∞ ∞ Z ∞ −st −st −st 2 e x¨ dt = xe˙ + s e x˙ dt = −u0 + s sX(s) − x0 = s X(s) − sx0 − u0. 0 0 0 The transformed ode is then

2 a s X − sx0 − u0 + b (sX − x0) + cX = G,

which is an easily solved linear algebraic equation for X = X(s). Finding the solution of the ode then requires taking the inverse Laplace transform of X = X(s) to obtain x = x(t).

97 98 LECTURE 30. LAPLACE TRANSFORM OF A CONSTANT-COEFFICIENT ODE

Problems for Lecture 30

1. Consider the inhomogeneous constant-coefﬁcient second-order ode given by

ax¨ + bx˙ + cx = g(t), x(0) = x0, x˙(0) = u0.

Determine the solution for X = X(s) in terms of the Laplace transform of g(t). Solutions to the Problems Lecture 31

Solution of an initial value problem View this lecture on YouTube

Example: Solve by Laplace transform methods x¨ + x = sin 2t with x(0) = 2 and x˙(0) = 1.

Taking the Laplace transform of both sides of the ode, we ﬁnd

2 s2X(s) − 2s − 1 + X(s) = , s2 + 4

where the Laplace transform of the second derivative and of sin 2t made use of line 16 and line 6a of the table in AppendixF. Solving for X(s), we obtain

2s + 1 2 X(s) = + . s2 + 1 (s2 + 1)(s2 + 4)

To determine the inverse Laplace transform from the table in AppendixF, we need to perform a partial fraction expansion of the second term:

2 as + b cs + d = + . (s2 + 1)(s2 + 4) s2 + 1 s2 + 4

By inspection, we can observe that a = c = 0 and that d = −b. A quick calculation shows that 3b = 2, or b = 2/3. Therefore,

2s + 1 2/3 2/3 X(s) = + − s2 + 1 s2 + 1 (s2 + 4) 2s 5/3 2/3 = + − . s2 + 1 s2 + 1 (s2 + 4)

From lines 6a and 7a of the table in AppendixF, we obtain the solution by taking inverse Laplace transforms of the three terms separately, where the values in the table are b = 1 in the ﬁrst two terms, and b = 2 in the third term: 5 1 x(t) = 2 cos t + sin t − sin 2t. 3 3

99 100 LECTURE 31. SOLUTION OF AN INITIAL VALUE PROBLEM

Problems for Lecture 31

1. Solve by Laplace transform methods x¨ + 5x˙ + 6x = e−t, with x(0) = 0 and x˙(0) = 0. Solutions to the Problems Practice quiz: The Laplace transform method 1. What is the Laplace transform of x(t) = e−t cos πt? s + 1 a) (s + 1)2 − π2 s + 1 b) (s + 1)2 + π2 s − 1 c) (s − 1)2 − π2 s − 1 d) (s − 1)2 + π2

2. If x¨ + x˙ − 6x = e−t, with x(0) = x˙(0) = 0, what is X(s)? 1 a) (s + 1)(s + 2)(s + 3) 1 b) (s − 1)(s + 2)(s + 3) 1 c) (s + 1)(s − 2)(s + 3) 1 d) (s + 1)(s + 2)(s − 3) 1 3. If X(s) = , what is x(t)? (s + 1)(s + 2)(s + 3) 1 1 a) e−t + e−2t + e−3t 2 2 1 1 b) − e−t + e−2t + e−3t 2 2 1 1 c) e−t − e−2t + e−3t 2 2 1 1 d) e−t + e−2t − e−3t 2 2 Solutions to the Practice quiz

101 102 LECTURE 31. SOLUTION OF AN INITIAL VALUE PROBLEM Lecture 32

The Heaviside step function View this lecture on YouTube

The Heaviside or unit step function, denoted here by uc(t), is zero for t < c and one for t ≥ c: ( 0, t < c; uc(t) = 1, t ≥ c.

The Heaviside function can be viewed as the step-up function. Using the Heaviside function both a step-down and a step-up, step-down function can also be deﬁned. The Laplace transform of the Heaviside function is determined by integration:

Z ∞ −cs −st e L{uc(t)} = e uc(t)dt = . 0 s

The Heaviside function can be used to represent a translation of a function f (t) a distance c in the positive t direction. We have ( 0, t < c; uc(t) f (t − c) = f(t-c), t ≥ c.

The Laplace transform is

Z ∞ −st −cs L{uc(t) f (t − c)} = e uc(t) f (t − c)dt = e F(s). 0

The translation of f (t) a distance c in the positive t direction corresponds to the multiplication of F(s) by the exponential e−cs. Piecewise-deﬁned inhomogeneous terms can be modeled using Heaviside functions. For example, consider the general case of a piecewise function deﬁned on two intervals: ( f (t), if t < c; f (t) = 1 f2(t), if t ≥ c.

Using the Heaviside function uc, the function f (t) can be written in a single line as

f (t) = f1(t) + f2(t) − f1(t) uc(t).

103 104 LECTURE 32. THE HEAVISIDE STEP FUNCTION

Problems for Lecture 32

1. Use the step-down function uc(t) to construct a step-up and a step-up, step-down function.

−cs 2. Prove that L{uc(t) f (t − c)} = e F(s).

3. Consider the piecewise continuous function given by  t, if t < 1; f (t) = 1, if t ≥ 1.

a) Express f (t) in a single line using the Heaviside function.

b) Find F = F(s). Solutions to the Problems Lecture 33

The Dirac delta function View this lecture on YouTube

The Dirac delta function, denoted as δ(t), is deﬁned by requiring that for any function f (t),

Z ∞ f (t)δ(t) dt = f (0). −∞

The usual view of the Dirac delta function is that it is zero everywhere except at t = 0, at which it is infinite in such a way that the integral is one. The Dirac delta function is technically not a function, but is what mathematicians call a distribution. Nevertheless, in most cases of practical interest, it can be treated like a function, where physical results are obtained following a final integration. There are several common ways to represent the Dirac delta function as a limit of a well-defined function. For our purposes, the most useful representation of the shifted Dirac delta function, δ(t − c), makes use of the step-up, step-down function constructed from Heaviside functions:

1 δ(t − c) = lim (uc−e(t) − uc+e(t)). e→0 2e

Before taking the limit, the well-deﬁned step-up, step-down function is zero except over a small interval of width 2e centered at t = c, over which it takes the large value 1/2e. The integral of this function is one, independent of the value of e. The Laplace transform of the Dirac delta function is easily found by integration using the deﬁnition of the delta function. With c > 0,

Z ∞ L{δ(t − c)} = e−stδ(t − c) dt = e−cs. 0

105 106 LECTURE 33. THE DIRAC DELTA FUNCTION

Problems for Lecture 33

Remember that the Dirac delta function is well-deﬁned only after it is integrated over. 1 1. Prove that δ(ax) = δ(x). |a| Z x 0 0 2. Show that uc(x) = δ(x − c) dx . −∞ d 3. Show that δ(x − c) = u (x) . dx c Solutions to the Problems Lecture 34

Solution of a discontinuous inhomogeneous term View this lecture on YouTube

Example: Solve x¨ + 3x˙ + 2x = 1 − u1(t), with x(0) = x˙(0) = 0

Here, the inhomogeneous term is a step-down function, from one to zero. Taking the Laplace transform of the ode using the table in AppendixF, we have

1 s2X(s) + 3sX(s) + 2X(s) = 1 − e−s , s with solution for X = X(s) given by

1 − e−s X(s) = . s(s + 1)(s + 2)

Deﬁning 1 F(s) = , s(s + 1)(s + 2) and using the table in AppendixF, the inverse Laplace transform of X(s) can be written as

x(t) = f (t) − u1(t) f (t − 1),

where f (t) is the inverse Laplace transform of F(s). To determine f (t), we need a partial fraction expansion of F(s), so we write

1 a b c = + + . s(s + 1)(s + 2) s s + 1 s + 2

Using the coverup method, we ﬁnd a = 1/2, b = −1, and c = 1/2, and from the table in AppendixF, determine

1 1 1 1 1 L−1 {F(s)} = L−1 − L−1 + L−1 2 s s + 1 2 s + 2 1 1 = − e−t + e−2t. 2 2

The full solution can then be written as

1 1 1 1 x(t) = − e−t + e−2t − u (t) − e−(t−1) + e−2(t−1) . 2 2 1 2 2

107 108 LECTURE 34. SOLUTION OF A DISCONTINUOUS INHOMOGENEOUS TERM

Problems for Lecture 34

1. Show that the solution in the lecture,

1 1 1 1 x(t) = − e−t + e−2t − u (t) − e−(t−1) + e−2(t−1) , 2 2 1 2 2 is continuous at t = 1.

2. Solve x¨ + x = 1 − u2π(t), with x(0) = 0 and x˙(0) = 0. Solutions to the Problems Lecture 35

Solution of an impulsive inhomogeneous term View this lecture on YouTube

Example: Solve x¨ + 3x˙ + 2x = δ(t) with x(0) = x˙(0) = 0. Assume the entire impulse occurs at t = 0+.

Taking the Laplace transform of the ode using the table in AppendixF, and applying the initial conditions, we have s2X + 3sX + 2X = 1.

Solving for X = X(s), we have 1 X(s) = , (s + 1)(s + 2) and a partial fraction expansion results in

1 1 1 = − . (s + 1)(s + 2) s + 1 s + 2

Taking the inverse Laplace transform using the table in AppendixF, we ﬁnd

x(t) = e−t − e−2t = e−t 1 − e−t .

Note that the function x = x(t) is continuous at t = 0, but that the ﬁrst derivative is not, since x˙(0−) = 0 and x˙(0+) = 1. Impulsive forces result in a discontinuity in the velocity.

109 110 LECTURE 35. SOLUTION OF AN IMPULSIVE INHOMOGENEOUS TERM

Problems for Lecture 35

1. Solve x¨ + x = δ(t) − δ(t − 2π), with x(0) = 0 and x˙(0) = 0. Solutions to the Problems Practice quiz: Discontinuous and impulsive inhomogeneous terms 1. The pictured function can be deﬁned using Heaviside step functions as

a) x(t) = −u1(t) + u2(t) + u3(t) − u4(t)

b) x(t) = −u1(t) + u2(t) − u3(t) + u4(t)

c) x(t) = u1(t) − u2(t) − u3(t) + u4(t)

d) x(t) = u1(t) − u2(t) + u3(t) − u4(t)

2. The solution to x¨ + x = 1 − u2π(t), with x(0) = 1 and x˙(0) = 0 is given by  1 − cos t, if t < 2π; a) x(t) = 0, if t ≥ 2π.  cos t, if t < 2π; b) x(t) = 1, if t ≥ 2π.  1 − sin t, if t < 2π; c) x(t) = 1, if t ≥ 2π.  1, if t < 2π; d) x(t) = cos t, if t ≥ 2π.

111 112 LECTURE 35. SOLUTION OF AN IMPULSIVE INHOMOGENEOUS TERM

3. The solution to x¨ + x = δ(t) − δ(t − 2π), with x(0) = 1 and x˙(0) = 0 is given by  cos t, if t < 2π; a) x(t) = 0, if t ≥ 2π.  sin t, if t < 2π; b) x(t) = 0, if t ≥ 2π.  cos t + sin t, if t < 2π; c) x(t) = cos t, if t ≥ 2π.  cos t + sin t, if t < 2π; d) x(t) = sin t, if t ≥ 2π. Solutions to the Practice quiz Lecture 36

The series solution method

View this lecture on YouTube

Example: Find the general solution of y00 + y = 0.

By now, you should know that the general solution is y(x) = a0 cos x + a1 sin x, with a0 and a1 constants. To ﬁnd a power series solution, we write

∞ n y(x) = ∑ anx ; n=0 and upon differentiating term-by-term

∞ ∞ 0 n−1 00 n−2 y (x) = ∑ nanx , y (x) = ∑ n(n − 1)anx . n=1 n=2

Substituting the power series for y and its derivatives into the differential equation, we obtain

∞ ∞ n−2 n ∑ n(n − 1)anx + ∑ anx = 0. (36.1) n=2 n=0

The power-series solution method requires combining the two sums into a single power series in x. We shift the summation index downward by two in the ﬁrst sum to obtain

∞ ∞ n−2 n ∑ n(n − 1)anx = ∑ (n + 2)(n + 1)an+2x . n=2 n=0

We can then combine the two sums in (36.1) to obtain

∞ n ∑ (n + 2)(n + 1)an+2 + an x = 0. n=0

For the equality to hold, the coefficient of each power of x must vanish separately. We therefore obtain the recurrence relation an a + = − , n = 0, 1, 2, . . . . n 2 (n + 2)(n + 1) We observe that even and odd coefficients decouple. We thus obtain two independent sequences starting with first term a0 or a1. Developing these sequences, we have for the first sequence,

1 1 1 a , a = − a , a = − a = a , 0 2 2 0 4 4 · 3 2 4! 0

113 114 LECTURE 36. THE SERIES SOLUTION METHOD and so on; and for the second sequence,

1 1 1 a , a = − a , a = − a = a , 1 3 3 · 2 1 5 5 · 4 3 5! 1 and so on. Using the principle of superposition, the general solution is therefore

x2 x4 x3 x5 y(x) = a 1 − + − ... + a x − + − ... 0 2! 4! 1 3! 5!

= a0 cos x + a1 sin x, as expected. 115

Problems for Lecture 36

1. Find two independent power series solutions of y00 − y = 0. Show that

y(x) = a0 cosh x + a1 sinh x, where ex + e−x ex − e−x cosh x = , sinh x = . 2 2 Solutions to the Problems 116 LECTURE 36. THE SERIES SOLUTION METHOD Lecture 37

Series solution of the Airy’s equation (Part A)

View this lecture on YouTube

We solve by series solution the Airy’s equation, an ode that arises in optics, ﬂuid mechanics, and quantum mechanics. Example: Find the general solution of y00 − xy = 0. Notice that there is a non-constant coefﬁcient.

We try the power-series ansatz ∞ n y(x) = ∑ anx , n=0

where the an coefﬁcients are to be determined. Finding the second derivative by differentiating term- by-term, the ode becomes ∞ ∞ n−2 n+1 ∑ n(n − 1)anx − ∑ anx = 0. (37.1) n=2 n=0 We shift the summation index of the ﬁrst sum down by three to obtain

∞ ∞ n−2 n+1 ∑ n(n − 1)anx = ∑ (n + 3)(n + 2)an+3x . n=2 n=−1

When combining the two sums in (37.1), we separate out the extra ﬁrst term in the ﬁrst sum. Therefore, (37.1) becomes the single power series

∞ n+1 2a2 + ∑ (n + 3)(n + 2)an+3 − an x = 0. n=0

Since all the coefﬁcients of powers of x equal zero, we ﬁnd a2 = 0 and obtain the recursion relation

1 a + = a . n 3 (n + 3)(n + 2) n

Three sequences of coefﬁcients—those starting with either a0, a1 or a2—decouple. Since a2 = 0, we ﬁnd immediately that a2 = a5 = a8 = a11 = ··· = 0. Starting with a0, we have

1 1 a , a = a , a = a , 0 3 3 · 2 0 6 6 · 5 · 3 · 2 0

117 118 LECTURE 37. SERIES SOLUTION OF THE AIRY’S EQUATION (PART A)

and so on; and starting with a1,

1 1 a , a = a , a = a , 1 4 4 · 3 1 7 7 · 6 · 4 · 3 1 and so on. The general solution for y = y(x), can therefore be written as

x3 x6 x4 x7 y(x) = a 1 + + + ... + a x + + + ... 0 3 · 2 6 · 5 · 3 · 2 1 4 · 3 7 · 6 · 4 · 3

= a0y0(x) + a1y1(x). 119

Problems for Lecture 37

1. Find two independent power series solutions of y00 + xy0 − y = 0. Keep terms up to x6. Solutions to the Problems 120 LECTURE 37. SERIES SOLUTION OF THE AIRY’S EQUATION (PART A) Lecture 38

Series solution of the Airy’s equation (Part B) View this lecture on YouTube

The Airy’s equation is given by y00 − xy = 0, and the general solution for y = y(x), is given by

x3 x6 x4 x7 y(x) = a 1 + + + ... + a x + + + ... 0 3 · 2 6 · 5 · 3 · 2 1 4 · 3 7 · 6 · 4 · 3

= a0y0(x) + a1y1(x).

Suppose we would like to graph the solutions y = y0(x) and y = y1(x) versus x by solving the differential equation y00 − xy = 0 numerically. What initial conditions should we use? Evidently, 0 y = y0(x) solves the ode with initial values y(0) = 1 and y (0) = 0, whereas y = y1(x) solves the ode with initial values y(0) = 0 and y0(0) = 1. The numerical solutions, obtained using Matlab, are shown in the ﬁgure below. Note that the solutions oscillate for negative x and grow exponentially for positive x. This can be understood by recalling that y00 + y = 0 has oscillatory sine and cosine solutions and y00 − y = 0 has exponential solutions.

Airy’s functions 2

1 0

y 0

−1

−2 −10 −8 −6 −4 −2 0 2

1 1

y 0

−1

−2 −10 −8 −6 −4 −2 0 2 x

Numerical solution of Airy’s equation

121 122 LECTURE 38. SERIES SOLUTION OF THE AIRY’S EQUATION (PART B)

Problems for Lecture 38

1. Given the two Airy’s functions y = y0(x) and y = y1(x) as deﬁned by their power series, solve the Airy’s equation y00 − xy = 0 with the initial conditions y(0) = 1 and y0(0) = 1. Solutions to the Problems Practice quiz: Series solutions 1. The value of cosh2 t − sinh2 t is equal to

a) −1

b)0

c)1

d)2

2. The general solution to y00 + x2y = 0 is given by

x3 x4 a) y(x) = a 1 + + ... + a x + + ... 0 12 1 20

x3 x4 b) y(x) = a 1 − + ... + a x − + ... 0 12 1 20

x4 x5 c) y(x) = a 1 + + ... + a x + + ... 0 12 1 20

x4 x5 d) y(x) = a 1 − + ... + a x − + ... 0 12 1 20 3. The general solution to y00 − xy0 + y = 0 is given by

x2 x4 a) y(x) = a 1 + + + ... + a x 0 2 24 1

x2 x4 b) y(x) = a 1 − + + ... + a x 0 2 24 1

x2 x4 c) y(x) = a 1 + − + ... + a x 0 2 24 1

x2 x4 d) y(x) = a 1 − − + ... + a x 0 2 24 1 Solutions to the Practice quiz

123 124 LECTURE 38. SERIES SOLUTION OF THE AIRY’S EQUATION (PART B) Week V

Systems of Differential Equations

125

127

We solve a coupled system of homogeneous ﬁrst-order differential equations with constant coef- ﬁcients. This system of odes can be written in matrix form, and we explain how to convert these equations into a standard matrix algebra eigenvalue problem. The two-dimensional solutions are visualized using phase portraits. We then discuss the important application of coupled harmonic oscillators and the calculation of normal modes. The normal modes are those motions for which the individual masses that make up the system oscillate with the same frequency. 128 Lecture 39

Systems of homogeneous linear ﬁrst-order odes View this lecture on YouTube

We consider a system of homogeneous linear odes with constant coefﬁcients given by

x˙1 = ax1 + bx2, x˙2 = cx1 + dx2, which can be written in matrix form as ! ! ! d x a b x 1 = 1 , dt x2 c d x2 or more succinctly as x˙ = Ax. We take as our ansatz x(t) = veλt, where v and λ are independent of t and v is a two-by-one column matrix. Upon substitution into the ode, we obtain

λveλt = Aveλt; and upon cancellation of the exponential, we obtain the eigenvalue problem

Av = λv.

The characteristic equation for our two-by-two matrix is given by

det (A − λI) = λ2 − (a + d)λ + (ad − bc) = 0.

We will demonstrate how to solve two separate cases: (i) eigenvalues of A are distinct and real; (ii) eigenvalues of A are complex conjugates. The third case for which the eigenvalues are repeated will be omitted. These three cases are analogous to those previously considered when solving the homogeneous constant-coefﬁcient second-order ode.

129 130 LECTURE 39. SYSTEMS OF LINEAR FIRST-ORDER ODES

Problems for Lecture 39

1. Consider the system of homogeneous linear odes with constant coefﬁcients given by

x˙1 = ax1 + cx2, x˙2 = cx1 + bx2.

Prove that the eigenvalues of the resulting characteristic equation are real. Solutions to the Problems Lecture 40

Distinct real eigenvalues View this lecture on YouTube

Example: Find the general solution of x˙1 = x1 + x2, x˙2 = 4x1 + x2. The equation in matrix form is given by ! ! ! d x 1 1 x 1 = 1 , dt x2 4 1 x2

or x˙ = Ax. With our ansatz x(t) = veλt, we obtain the eigenvalue problem Av = λv. The characteristic equation of A is given by

det (A − λI) = λ2 − 2λ − 3 = (λ − 3)(λ + 1) = 0,

and the eigenvalues are found to be λ1 = −1 and λ2 = 3. To determine the corresponding eigenvectors, we substitute the eigenvalues successively into (A − λI)v = 0, and denote the eigenvectors as T T v1 = v11 v21 and v2 = v12 v22 . For λ1 = −1, and unknown eigenvector v1, we obtain from (A − λ1I)v1 = 0 the result v21 = −2v11. Recall that an eigenvector is only unique up to multiplication by a constant: we may therefore take

v11 = 1. For λ2 = 3, and unknown eigenvector v2, we obtain from (A − λ2I)v2 = 0 the result v22 = 2v12. Here, we take v12 = 1. Therefore, our eigenvalues and eigenvectors are given by ! ! 1 1 λ1 = −1, v1 = ; λ2 = 3, v2 = . −2 2

Using the principle of superposition, the general solution to the system of odes is given by

λ1t λ2t x(t) = c1v1e + c2v2e ;

or explicitly writing out the components,

−t 3t −t 3t x1(t) = c1e + c2e , x2(t) = −2c1e + 2c2e .

131 132 LECTURE 40. DISTINCT REAL EIGENVALUES

Problems for Lecture 40

1. Using matrix algebra methods, ﬁnd the general solution of x˙1 = −x2 and x˙2 = −2x1 − x2. Solutions to the Problems Lecture 41

Complex-conjugate eigenvalues View this lecture on YouTube 1 1 Example: Find the general solution of x˙ = − x + x and x˙ = −x − x . 1 2 1 2 2 1 2 2 The equations in matrix form are ! ! ! d x − 1 1 x 1 = 2 1 . dt 1 x2 −1 − 2 x2

The ansatz x = veλt leads to the characteristic equation

5 det (A − λI) = λ2 + λ + = 0, 4

which has complex-conjugate roots. We denote the two eigenvalues as

1 1 λ = − + i and λ¯ = − − i. 2 2

The eigenvectors also occur as a complex-conjugate pair, and the eigenvector v associated with the T eigenvalue λ satisﬁes −iv1 + v2 = 0. Normalizing with v1 = 1, we have v = 1 i . We have therefore determined two independent complex solutions to the system of odes, that is,

¯ veλt and ¯veλt,

and we can form a linear combination of these two complex solutions to construct two independent real solutions. Namely, the two real solutions are Re{veλt} and Im{veλt}. We have ( ! ) ! λt 1 (− 1 +i)t −t/2 cos t Re{ve } = Re e 2 = e ; i − sin t

and ( ! ) ! λt 1 (− 1 +i)t −t/2 sin t Im{ve } = Im e 2 = e . i cos t

Taking a linear superposition of these two real solutions yields the general solution to the system of odes, given by ! !! cos t sin t x = e−t/2 A + B ; − sin t cos t

or explicitly writing out the components,

−t/2 −t/2 x1 = e (A cos t + B sin t) , x2 = e (−A sin t + B cos t) .

133 134 LECTURE 41. COMPLEX-CONJUGATE EIGENVALUES

Problems for Lecture 41

1. Using matrix algebra methods, ﬁnd the general solution of x˙1 = x1 − 2x2 and x˙2 = x1 + x2. Solutions to the Problems Practice quiz: Systems of differential equations

1. The system of odes given by x˙1 = ax1 + bx2 and x˙2 = cx1 + dx2 has the same characteristic equation as the second order ode given by

a) ax¨ + bx˙ + cx = 0

b) x¨ + (a + b)x˙ + (c + d)x = 0

c) x¨ − (a + d)x˙ + (ad − bc)x = 0

d) (a + b)x¨ + (b + c)x˙ + (c + d)x = 0

2. The general solution of x˙1 = x1 + 2x2 and x˙2 = 2x1 + x2 is given by

3t −t a) x1 = c1e + c2e 3t −t x2 = c1e + c2e

3t −t b) x1 = c1e + c2e 3t −t x2 = c1e − c2e

3t −t c) x1 = c1e + c2e −t 3t x2 = c1e + c2e

3t −t d) x1 = c1e + c2e −t 3t x2 = c1e − c2e

3. The general solution of x˙1 = −2x1 + x2 and x˙2 = −x1 − 2x2 is given by

−t a) x1 = e (A cos 2t + B sin 2t) −t x2 = e (A sin 2t + B cos 2t)

−2t b) x1 = e (A cos t + B sin t) −2t x2 = e (A sin t + B cos t)

−t c) x1 = e (A cos 2t + B sin 2t) −t x2 = e (−A sin 2t + B cos 2t)

−2t d) x1 = e (A cos t + B sin t) −2t x2 = e (−A sin t + B cos t) Solutions to the Practice quiz

135 136 LECTURE 41. COMPLEX-CONJUGATE EIGENVALUES Lecture 42

Phase portraits View this lecture on YouTube

The solution of a system of two first-order odes for x1 and x2 can be visualized by drawing a phase portrait, with “x-axis” x1 and “y-axis” x2. Each curve drawn on the phase portrait corresponds to a different initial condition, and can be viewed as the trajectory of a particle at position (x1, x2) moving with a velocity given by (x˙1, x˙2). For a two-by-two system, written as x˙ = Ax, the point x = (0, 0) is called an equilibrium point, or fixed point, of the system. If x is at the fixed point initially, then x remains there for all time because x˙ = 0 at the fixed point. This fixed point, or equilibrium, may be stable or unstable, and the qualitative picture of the phase portrait depends on the stability of the equilibrium, which in turn depends on the eigenvalues of the characteristic equation. If there are two distinct real eigenvalues of the same sign, we say that the fixed point is a node. When the eigenvalues are both negative the fixed point is a stable node, and when they are both positive, the fixed point is an unstable node. If the eigenvalues have opposite sign, the fixed point is a saddle point. If there are complex-conjugate eigenvalues, we say that the fixed point is a spiral. A spiral can be stable or unstable and this depends on the sign of the real part of the eigenvalues. If the real part is negative, then the solution decays exponentially and the fixed point corresponds to a stable spiral; if the real part is positive then the solution grows exponentially and the fixed point corresponds to an unstable spiral. Furthermore, a spiral may wind around the fixed point clockwise or counterclockwise, and this so-called handedness of the spiral can be determined by examining the differential equations directly. In the next few lectures, we will present phase portraits representing nodes, saddle points and spirals.

137 138 LECTURE 42. PHASE PORTRAITS

Problems for Lecture 42

1. Determine if the ﬁxed points of the following systems are nodes, saddle points, or spirals, and determine their stability:

a) √ √ x˙1 = −3x1 + 2x2, x˙2 = 2x1 − 2x2;

x˙1 = x1 + x2, x˙2 = 4x1 + x2;

c) 1 1 x˙ = − x + x , x˙ = −x − x . 1 2 1 2 2 1 2 2 Solutions to the Problems Lecture 43

Stable and unstable nodes View this lecture on YouTube

When there are two distinct real eigenvalues of the same sign, the ﬁxed point is called a node. Consider the differential equations given by √ √ x˙1 = −3x1 + 2x2, x˙2 = 2x1 − 2x2.

An eigenvalue analysis of this system results in eigenvalues and eigenvectors given by ! ! 1 1 λ1 = −4, v1 = √ ; λ2 = −1, v2 = √ ; − 2/2 2

and the general solution of the system is written as

λ1t λ2t x(t) = c1v1e + c2v2e .

Because λ1, λ2 < 0, both exponential solutions for x = x(t) decay in time and x → (0, 0) as t → ∞. This is why we say that the node is stable. If both eigenvalues were positive, the node would be unstable. To draw the phase portrait, one considers the eigenfunctions. On the one hand, suppose initial conditions are such that c2 = 0. The solution x(t) is then equal to a scalar function of time times the eigenvector v . Since x is proportional to v , trajectories with c = 0 must lie on the line x = √ 1 1 2 2 − 2x1/2. This line can be drawn on the phase portrait with arrow pointing in towards the origin.

2 On the other hand, suppose initial conditions are

such that c1 = 0. The solution is then equal to a scalar times the eigenvector v . Since x is proportional to v , 1 2 √ 2 trajectories with c1 = 0 must lie on the line x2 = 2x1. 2

This line can also be drawn on the phase portrait with x 0 arrow again pointing in towards the origin.

Because |λ1| > |λ2|, the trajectory approaches the -1 ﬁxed point quicker along v1 than along v2, resulting in a ﬁnal approach to the origin along v2, as is evident -2 in the drawn phase portrait. -2 -1 0 1 2 x 1

139 140 LECTURE 43. STABLE AND UNSTABLE NODES

Problems for Lecture 43

1. Consider the system of differential equations given by

x˙1 = 2x1 + x2, x˙2 = x1 + 2x2.

Determine the eigenvalues and eigenvectors, and sketch the phase portrait. Solutions to the Problems Lecture 44

Saddle points View this lecture on YouTube

If there are two distinct real eigenvalues of opposite sign, the ﬁxed point is a saddle point. Consider the differential equations given by

x˙1 = x1 + x2, x˙2 = 4x1 + x2.

An eigenvalue analysis of this system results in eigenvalues and eigenvectors given by ! ! 1 1 λ1 = −1, v1 = ; λ2 = 3, v2 = ; −2 2 and the general solution is again written as

λ1t λ2t x(t) = c1v1e + c2v2e .

Because λ1 < 0, trajectories approach the fixed point along the direction of the first eigenvector, and because λ2 > 0, trajectories move away from the fixed point along the direction of the second eigenvector. Ultimately, a saddle point is an unstable equilibrium because for any initial conditions such that c2 6= 0, |x(t)| → ∞ as t → ∞. To draw a phase portrait of the solution, we first consider motion along the directions 2 of the eigenvectors. Along the first eigenvector, we have x2 = −2x1, and this line is to be drawn on the phase portrait with arrow 1 pointing in towards origin. Along the second eigenvector, we have x2 = 2x1, and this 2 line is to be drawn on the phase portrait with x 0 arrow pointing away from the origin. Imme- diately, we see that the motion is in towards -1 the origin in the direction of the first eigenvector, and away from the origin in the direction of the second eigenvector. The remainder of -2 the phase trajectories can be sketched or com- -2 -1 0 1 2 x puter drawn, and the resulting phase portrait 1 is shown.

141 142 LECTURE 44. SADDLE POINTS

Problems for Lecture 44

1. Consider the system of differential equations given by

x˙1 = −x1 + 3x2, x˙2 = 2x1 + 4x2.

Determine the eigenvalues and eigenvectors, and sketch the phase portrait. Solutions to the Problems Lecture 45

Spiral points View this lecture on YouTube

If there are complex conjugate eigenvalues, the ﬁxed point is a spiral point. Consider the system of differential equations given by

1 1 x˙ = − x + x , x˙ = −x − x . 1 2 1 2 2 1 2 2

An eigenvalue analysis of this system results in the complex eigenvalue and eigenvector ! 1 1 λ = − + i, v = , 2 i

and their complex conjugates. The general solution is written as " ! !# cos t sin t x(t) = e−t/2 A + B . − sin t cos t

The trajectories in the phase portrait are spirals centered at the ﬁxed point. If the Re{λ} > 0 the trajectories spiral out, and if Re{λ} < 0 they spiral in. The spirals around the ﬁxed point may be clockwise or counterclockwise, depending on the governing equations.

2 Here, since Re{λ} = −1/2 < 0, the trajectories spiral into the origin. To determine whether the spiral is clockwise 1 or counterclockwise, we can examine the

time derivatives at the point (x1, x2) = 2

x 0 (0, 1). At this point in the phase space,

(x˙1, x˙2) = (1, −1/2), and a particle on this trajectory moves to the right and -1 downward, indicating a clockwise spiral. The corresponding phase portrait is -2 shown. -2 -1 0 1 2 x 1

143 144 LECTURE 45. SPIRAL POINTS

Problems for Lecture 45

1. Consider the system of differential equations given by

x˙1 = x1 + x2, x˙2 = −x1 + x2.

Determine the eigenvalues and eigenvectors, and sketch the phase portrait. Solutions to the Problems Practice quiz: Phase portraits 1. Select the correct phase portrait for the following system of differential equations:

x˙1 = 2x1 + x2, x˙2 = x1 + 2x2.

2 2

1 1 2 2

x 0 x 0

-1 -1

-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 x x (a) 1 (b) 1

2 2

1 1 2 2

x 0 x 0

-1 -1

-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 x x (c) 1 (d) 1

2. Select the correct phase portrait for the following system of differential equations:

x˙1 = −x1 + 3x2, x˙2 = 2x1 + 4x2.

(a) (b)

145 146 LECTURE 45. SPIRAL POINTS

3. Select the correct phase portrait for the following system of differential equations:

x˙1 = x1 + x2, x˙2 = −x1 + x2.

2 2

1 1 2 2

x 0 x 0

-1 -1

-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 x x (a) 1 (b) 1

2 2

1 1 2 2

x 0 x 0

-1 -1

-2 -2 -2 -1 0 1 2 -2 -1 0 1 2 x x (c) 1 (d) 1 Solutions to the Practice quiz Lecture 46

Coupled oscillators View this lecture on YouTube

Coupled harmonic oscillators: two masses, three springs

The normal modes of a physical system are oscillations of the entire system that take place at a single frequency. Perhaps the simplest example of a system containing two distinct normal modes is the coupled mass-spring system shown above. We will see that an eigenvector analysis of this system, similar to what we have just done for coupled ﬁrst-order odes, will exhibit the normal modes and reveal the true nature of the oscillation.

In the ﬁgure, the position variables x1 and x2 are measured from the equilibrium positions of the masses. Hooke’s law states that the spring force is linearly proportional to the extension length of the spring, measured from equilibrium. By considering the extension of each spring and the sign of each force, Newton’s law F = ma written separately for each mass is

mx¨1 = −kx1 − K(x1 − x2),

mx¨2 = −kx2 − K(x2 − x1).

Collecting terms proportional to x1 and x2 results in

mx¨1 = −(k + K)x1 + Kx2,

mx¨2 = Kx1 − (k + K)x2.

The equations for the coupled mass-spring system form a system of two homogeneous linear second- order odes. In matrix form, mx¨ = Ax, or explicitly, ! ! ! d2 x −(k + K) K x 1 = 1 m 2 . dt x2 K −(k + K) x2

147 148 LECTURE 46. COUPLED OSCILLATORS

Problems for Lecture 46

k k m m

x1 x2

Coupled harmonic oscillators: two masses, two springs

1. Consider the mass-spring system shown above. Determine the matrix equation represented as mx¨ = Ax. Solutions to the Problems Lecture 47

Normal modes (eigenvalues) View this lecture on YouTube We now solve the two masses, three springs system with governing equation ! ! ! d2 x −(k + K) K x 1 = 1 m 2 . dt x2 K −(k + K) x2

We try the ansatz x = vert, and obtain the eigenvalue problem Av = λv, with λ = mr2. The eigenvalues are determined by solving the characteristic equation

det (A − λI) = (λ + k + K)2 − K2 = 0;

and the two solutions for λ are given by

λ1 = −k, λ2 = −(k + 2K). √ The corresponding values of r in our ansatz x = vert, with r = ± λ/m, are

√ q r1 = i k/m, r¯1, r2 = i (k + 2K)/m, r¯2.

Since the values of r are pure imaginary, we know that x1(t) and x2(t) will oscillate with angular frequencies ω1 = Im{r1} and ω2 = Im{r2}, that is, √ q ω1 = k/m, ω2 = (k + 2K)/m.

These are the so-called frequencies of the two normal modes. A normal mode with angular frequency

ωi is periodic with period Ti = 2π/ωi. In the next lecture, we ﬁnd the eigenvectors and physically intepret the results.

149 150 LECTURE 47. NORMAL MODES (EIGENVALUES) Lecture 48

Normal modes (eigenvectors) View this lecture on YouTube We are solving the two masses, three springs system with governing equation ! ! ! d2 x −(k + K) K x 1 = 1 m 2 . dt x2 K −(k + K) x2

We tried the ansatz x = vert, and obtained the eigenvalue problem Av = λv, with λ = mr2. The eigenvalues were found to be

λ1 = −k, λ2 = −(k + 2K).

The eigenvectors, or so-called normal modes of oscillations, correspond to the mass-spring system

oscillating at a single frequency. The eigenvector with eigenvalue λ1 satisﬁes

−Kv11 + Kv12 = 0, √ so that v11 = v12. The normal mode with frequency ω1 = k/m thus follows a motion where x1 = x2. During this motion the center spring length does not change, and the frequency is independent of K.

Next, we determine the eigenvector with eigenvalue λ2:

Kv21 + Kv22 = 0,

p so that v21 = −v22. The normal mode with frequency ω2 = (k + 2K)/m thus follows a motion where x1 = −x2. During this motion the two equal masses symmetrically push or pull against each side of the middle spring. A general solution for x(t) can be constructed from the eigenvalues and eigenvectors. Our ansatz was x = vert, and for each of two eigenvectors v, we have a pair of complex conjugate values for r. Accordingly, we ﬁrst apply the principle of superposition to obtain four real solutions, and then apply √ p the principle again to obtain the general solution. With ω1 = k/m and ω2 = (k + 2K)/m, the general solution is given by ! ! ! x1 1 1 = (A cos ω1t + B sin ω1t) + (C cos ω2t + D sin ω2t) , x2 1 −1 where the now real constants A, B, C, and D can be determined from the four independent initial conditions, x1(0), x2(0), x˙1(0), and x˙2(0).

151 152 LECTURE 48. NORMAL MODES (EIGENVECTORS)

Problems for Lecture 48

k k m m

x1 x2

Coupled harmonic oscillators: two masses, two springs

1. Consider the matrix equation obtained from the mass-spring system shown above. Find the angular frequencies and the eigenvectors of the two normal modes. Solutions to the Problems Practice quiz: Normal modes 1. The matrix equation for the pictured mass-spring system is given by ! ! ! d2 x −k k x 1 = 1 a) m 2 dt x2 k −k x2 ! ! ! d2 x −2k k x 1 = 1 b) m 2 dt x2 k −k x2 ! ! ! d2 x −3k k x 1 = 1 c) m 2 dt x2 k −k x2 ! ! ! d2 x −4k k x 1 = 1 d) m 2 dt x2 k −k x2 2. The angular frequencies of the normal modes for the mass-spring system of Question 1 are approximately given by r r k k a) ω = 0.77 , ω = 1.85 1 m 2 m r r k k b) ω = 0.84 , ω = 2.14 1 m 2 m r r k k c) ω = 1.45 , ω = 2.34 1 m 2 m r r k k d) ω = 1.82 , ω = 2.66 1 m 2 m 3. The eigenvectors of the normal modes for the mass-spring system of Question 1 are approximately given by ! ! 1 1 a)v 1 = , v2 = 1.32 −0.22 ! ! 1 1 b)v 1 = , v2 = 1.98 −0.35 ! ! 1 1 c)v 1 = , v2 = 2.41 −0.41 ! ! 1 1 d)v 1 = , v2 = 2.92 −0.63 Solutions to the Practice quiz

153 154 LECTURE 48. NORMAL MODES (EIGENVECTORS) Week VI

Partial Differential Equations

155

157

To solve a partial differential equation, we must first define the Fourier series, and the Fourier sine and cosine series. We then derive the one-dimensional diffusion equation, which is a partial differential equation for the time-evolution of the concentration of a dye over one spatial dimension. We proceed to solve this equation using the method of separation of variables. This method yields a mathematical solution for a dye diffusing length-wise within a finite length pipe. 158 Lecture 49

Fourier series View this lecture on YouTube

Fourier series will be needed to solve the diffusion equation. A periodic function f (x) with period 2L can be represented as a Fourier series in the form

∞ a0 nπx nπx f (x) = + ∑ an cos + bn sin . (49.1) 2 n=1 L L

Determination of the coefficients a0, a1, a2, . . . and b1, b2, b3, . . . makes use of orthogonality relations for sine and cosine. We first define the Kronecker delta δnm as ( 1, if n = m; δnm = 0, otherwise.

The orthogonality relations for n and m positive integers are then given as the integration formulas

Z L mπx nπx Z L mπx nπx cos cos dx = Lδnm, sin sin dx = Lδnm, −L L L −L L L Z L mπx nπx cos sin dx = 0. −L L L

To determine the coefﬁcient an, we multiply both sides of (49.1) by cos (nπx/L), change the dummy summation variable from n to m, and integrate over x from −L to L. To determine the coefﬁcient

bn, we do the same except multiply by sin (nπx/L). To illustrate the procedure, when ﬁnding bn, integration will result in

Z L nπx ∞ Z L nπx mπx f (x) sin dx = ∑ bm sin sin dx −L L m=1 −L L L ∞ = ∑ bm (Lδnm) m=1

= Lbn,

which can be solved for bn. The ﬁnal results for the coefﬁcients are given by

1 Z L nπx 1 Z L nπx an = f (x) cos dx, bn = f (x) sin dx. L −L L L −L L

159 160 LECTURE 49. FOURIER SERIES

Problems for Lecture 49

1. Let ∞ a0 nπx nπx f (x) = + ∑ an cos + bn sin . 2 n=1 L L a) Show that f (x + 2L) = f (x), that is, f (x) is a periodic function with period 2L.

b) Show that a0 is twice the average value of f (x). Solutions to the Problems Lecture 50

Fourier sine and cosine series View this lecture on YouTube

We now know that the Fourier series of a periodic function with period 2L is given by

∞ a0 nπx nπx f (x) = + ∑ an cos + bn sin , 2 n=1 L L

with 1 Z L nπx 1 Z L nπx an = f (x) cos dx, bn = f (x) sin dx. L −L L L −L L The Fourier series simpliﬁes if f (x) is an even function such that f (−x) = f (x), or an odd function such that f (−x) = − f (x). We make use of the following facts. The function cos (nπx/L) is an even function and sin (nπx/L) is an odd function. The product of two even functions is an even function. The product of two odd functions is an even function. The product of an even and an odd function is an odd function. And if g(x) is an even function, then

Z L Z L g(x) dx = 2 g(x) dx; −L 0

and if g(x) is an odd function, then Z L g(x) dx = 0. −L Regarding the Fourier series, then, if f (x) is even we have

2 Z L nπx an = f (x) cos dx, bn = 0; L 0 L

and the Fourier series for an even function is given by the Fourier cosine series

∞ a0 nπx f (x) = + ∑ an cos , f (x) even. 2 n=1 L

If f (x) is odd, then 2 Z L nπx an = 0, bn = f (x) sin dx; L 0 L and the Fourier series for an odd function is given by the Fourier sine series

∞ nπx f (x) = ∑ bn sin , f (x) odd. n=1 L

161 162 LECTURE 50. FOURIER SINE AND COSINE SERIES

Problems for Lecture 50

1. Show that the Fourier series for an odd function satisﬁes f (0) = 0 and that the Fourier series for an even function satisﬁes f 0(0) = 0. Solutions to the Problems Lecture 51

Fourier series: example View this lecture on YouTube

Example: Determine the Fourier series of the triangle wave, shown in the following ﬁgure:

-1 -2π -π 0 π 2π

The triangle wave

Evidently, the triangle wave is an even function of x with period 2π, and its deﬁnition over half a period is 2x f (x) = 1 − , 0 < x < π. π Because f (x) is even, it can be represented by a Fourier cosine series (with L = π) given by

∞ Z π a0 2 f (x) = + ∑ an cos nx, with an = f (x) cos nx dx. 2 n=1 π 0

The coefﬁcient a0/2 is the average value of f (x) over one period, which is clearly zero. The coefﬁcients an for n > 0 are

( 2 2 2 Z π 2x 4 8/(n π ), if n odd; an = 1 − cos nx dx = 1 − cos nπ = π 0 π n2π2 0, if n even.

The Fourier cosine series for the triangle wave is therefore given by

8 cos 3x cos 5x f (x) = cos x + + + ... . π2 32 52

Convergence of this series is rapid. As an interesting aside, evaluation of this series at x = 0, using f (0) = 1, yields in an inﬁnite series for π2, which is usually written in the form

π2 1 1 = 1 + + + .... 8 32 52

163 164 LECTURE 51. FOURIER SERIES: EXAMPLE

Problems for Lecture 51

1. Determine the Fourier series of the square wave, shown in the following ﬁgure:

-1 -2π -π 0 π 2π

The square wave

Solutions to the Problems Practice quiz: Fourier series 1. The Fourier series of a periodic function of period 2L is given by ∞ a0 nπx nπx f (x) = + ∑ an cos + bn sin . The value of f (0) is given by 2 n=1 L L a a) 0 2 ∞ a0 b) + ∑ an 2 n=1 ∞ c) ∑ bn n=1 ∞ a0 d) + ∑ (an + bn) 2 n=1 2. The Fourier series of the square wave in the shown ﬁgure is given by

4 cos 3x cos 5x a) f (x) = cos x + + + ... π 3 5 4 cos 3x cos 5x b) f (x) = cos x − + − ... π 3 5 4 sin 3x sin 5x c) f (x) = sin x + + + ... π 3 5 4 sin 3x sin 5x d) f (x) = sin x − + − ... π 3 5

3. The Fourier series of the square wave in the shown ﬁgure is given by

4 cos 3x cos 5x a) g(x) = cos x + + + ... π 3 5 4 cos 3x cos 5x b) g(x) = cos x − + − ... π 3 5 4 sin 3x sin 5x c) g(x) = sin x + + + ... π 3 5 4 sin 3x sin 5x d) g(x) = sin x − + − ... π 3 5

Solutions to the Practice quiz

165 166 LECTURE 51. FOURIER SERIES: EXAMPLE Lecture 52

The diffusion equation View this lecture on YouTube To derive the diffusion equation in one spacial dimension, we imagine a still liquid in a pipe of constant cross sectional area. A small quantity of dye is placed uniformly across a cross section of the pipe and allowed to diffuse up and down the pipe. We define u(x, t) to be the concentration (mass per unit length) of the dye at position x along the pipe, and our goal is to find the equation satisfied by u.

The mass of dye at time t in a pipe volume located between x and x + ∆x is given to order ∆x by

M = u(x, t)∆x.

This mass of dye diffuses in and out of the cross sectional ends as shown above. We assume the rate of diffusion is proportional to the concentration gradient, a relationship known as Fick’s law of diffusion. Fick’s law assumes the mass ﬂux J (mass per unit time) across a cross section of the pipe is given by

J = −Dux, where D > 0 is the diffusion constant, and we have used the notation ux = ∂u/∂x. The mass flux is opposite in sign to the gradient of concentration so that the flux is from high concentration to low concentration. The time rate of change in the mass of dye is given by the difference between the mass flux in and the mass flux out of the cross sectional volume. A positive mass flux signifies diffusion from left to right. Therefore, the time rate of change of the dye mass is given by

dM = J(x, t) − J(x + ∆x, t); dt or rewriting in terms of u(x, t),

ut(x, t)∆x = −Dux(x, t) + Dux(x + ∆x, t) = D (ux(x + ∆x, t) − ux(x, t)) .

Dividing by ∆x and taking the limit ∆x → 0 results in the diffusion equation, given by

ut = Duxx.

167 168 LECTURE 52. THE DIFFUSION EQUATION

Problems for Lecture 52

1. Consider the one-dimensional diffusion equation in a pipe of length L. Nondimensionalize the diffusion equation using L as the unit of length and L2/D as the unit of time. Solutions to the Problems Lecture 53

Solution of the diffusion equation (separation of variables) View this lecture on YouTube

We begin to solve the diffusion equation

ut = Duxx

for the concentration u(x, t) in a pipe of length L. The solution method we use is called separation of variables. We assume that u(x, t) can be written as a product of two other functions, one dependent only on position x and the other dependent only on time t. That is, we make the ansatz

u(x, t) = X(x)T(t).

We will ﬁnd all possible solutions of this type—there will in fact be an inﬁnite number—and then use the principle of superposition to combine them to construct a general solution. Substitution of our ansatz into the diffusion equation results in

XT0 = DX00T,

which we rewrite by separating the x and t dependence to opposite sides of the equation:

X00 1 T0 = . X D T

The left-hand side of this equation is independent of t and the right-hand side is independent of x. Both sides of this equation are therefore independent of both t and x and must be equal to a constant, which we call −λ. Introduction of this separation constant results in the two ordinary differential equations X00 + λX = 0, T0 + λDT = 0. (53.1)

To proceed further, boundary conditions need to be speciﬁed at the pipe ends. We will assume here that the ends open up into large reservoirs of clear ﬂuid so that u(0, t) = u(L, t) = 0 for all times. Applying these boundary conditions to our ansatz results in

u(0, t) = X(0)T(t) = 0, u(L, t) = X(L)T(t) = 0.

Since these boundary conditions are valid for all t, we must have X(0) = X(L) = 0. These are called homogeneous Dirichlet boundary conditions. In the next lecture, we proceed to solve for X = X(x).

169 170 LECTURE 53. SOLUTION OF THE DIFFUSION EQUATION (SEPARATION OF VARIABLES)

Problems for Lecture 53

1. If the ends of the pipe were closed preventing a further diffusion of dye, then the mass ﬂux through

the pipe’s ends would be zero, and the appropriate boundary conditions would be ux(0, t) = ux(L, t) = 0 for all times. Determine the appropriate boundary conditions on X = X(x). Solutions to the Problems Lecture 54

Solution of the diffusion equation (eigenvalues) View this lecture on YouTube

We now solve the equation for X = X(x) with homogeneous Dirichlet boundary conditions:

X00 + λX = 0, X(0) = X(L) = 0.

Clearly, the trivial solution X(x) = 0 is a solution, and we will see that nontrivial solutions exist only for discrete values of λ. These discrete values and the corresponding functions X = X(x) are called the eigenvalues and eigenfunctions of the ode. The form of the general solution of the ode depends on the sign of λ. One can show that nontrivial solutions exist only when λ > 0. We therefore write λ = µ2, and determine the general solution of

X00 + µ2X = 0

to be X(x) = A cos µx + B sin µx.

The boundary condition at x = 0 results in A = 0, and the boundary condition at x = L results in

B sin µL = 0.

The solution B = 0 results in the unsought trivial solution. Therefore, we must have

sin µL = 0,

which is an equation for µ, with solutions given by µn = nπ/L, where n is a nonzero integer. We have thus determined all the unique eigenvalues λ = µ2 > 0 to be

2 λn = (nπ/L) , n = 1, 2, 3, . . . , with corresponding eigenfunctions

Xn = sin (nπx/L).

There is no need here to include the multiplication by an arbitrary constant, which will be added later.

171 172 LECTURE 54. SOLUTION OF THE DIFFUSION EQUATION (EIGENVALUES)

Problems for Lecture 54

1. Show that the equations X00 + λX = 0, X(0) = X(L) = 0

have no nontrivial solutions for λ ≤ 0.

2. Solve the following equation for X = X(x) with given boundary conditions:

X00 + λX = 0, X0(0) = X0(L) = 0.

Solutions to the Problems Practice quiz: Separable partial differential equations

1. The units of the diffusion constant D in the diffusion equation ut = Duxx are

a) lt2

b) lt−2

c) l2t

d) l2t−1

2 2. The one-dimensional wave equation is given by utt = c uxx. With u(x, t) = X(x)T(t), the separated ordinary differential equations can be written as

a) X0 + λX = 0, T0 + λc2T = 0

b) X00 + λX = 0, T0 + λc2T = 0

c) X0 + λX = 0, T00 + λc2T = 0

d) X00 + λX = 0, T00 + λc2T = 0

3. The eigenvalues and eigenfunctions of the differential equation X00 + λX = 0, with mixed boundary conditions X(0) = 0 and X0(L) = 0, are given by nπ 2 nπx a) λ = , X = sin , n = 1, 2, 3, . . . n L n L nπ 2 nπx b) λ = , X = cos , n = 0, 1, 2, 3, . . . n L n L (2n − 1)π 2 (2n − 1)πx c) λ = , X = sin , n = 1, 2, 3, . . . n 2L n 2L

d) λ0 = 0, X0 = 1, (2n − 1)π 2 (2n − 1)πx λ = , X = cos , n = 1, 2, 3, . . . n 2L n 2L Solutions to the Practice quiz

173 174 LECTURE 54. SOLUTION OF THE DIFFUSION EQUATION (EIGENVALUES) Lecture 55

Solution of the diffusion equation (Fourier series) View this lecture on YouTube

2 With the eigenvalues λn = (nπ/L) , the differential equation for T = T(t) becomes T0 + n2π2D/L2 T = 0, which has solution proportional to −n2π2Dt/L2 Tn = e .

Therefore, with ansatz u(x, t) = X(x)T(t) and eigenvalues λn, we conclude that the functions

−n2π2Dt/L2 un(x, t) = sin (nπx/L)e satisfy the diffusion equation and the spatial boundary conditions for every positive integer n. The principle of linear superposition for homogeneous linear differential equations then states that the general solution to the diffusion equation with the spatial boundary conditions is given by

∞ ∞ −n2π2Dt/L2 u(x, t) = ∑ bnun(x, t) = ∑ bn sin (nπx/L)e . n=1 n=1

The final solution step is to determine the unknown coefficients bn by satisfying conditions on the initial dye concentration. We assume that u(x, 0) = f (x), where f (x) is some specific function defined on 0 ≤ x ≤ L. At t = 0, we have ∞ f (x) = ∑ bn sin (nπx/L). n=1 We immediately recognize this equation as a Fourier sine series for an odd function f (x) with period 2L. A Fourier sine series results because of the boundary condition f (0) = 0. From our previous solution for the coefficients of a Fourier sine series, we determine that

2 Z L nπx bn = f (x) sin dx. L 0 L

175 176 LECTURE 55. SOLUTION OF THE DIFFUSION EQUATION (FOURIER SERIES)

Problems for Lecture 55

1. Determine the solution to the diffusion equation ut = Duxx for the concentration u = u(x, t) with closed pipe ends corresponding to the boundary conditions ux(0) = ux(L) = 0, and with general initial conditions given by u(x, 0) = f (x). Use the results of the previously solved problems.

Solutions to the Problems Lecture 56

Diffusion equation: example View this lecture on YouTube

Example: Determine the concentration of a dye in a pipe of length L, where the dye is initially concentrated

uniformly across the center of the pipe with total mass M0, and the ends of the pipe open onto a large reservoir.

We solve the diffusion equation with homogeneous Dirichlet boundary conditions, and model the

initial concentration of the dye by a delta-function centered at x = L/2, that is, u(x, 0) = M0δ(x − L/2). The Fourier sine series coefﬁcients are therefore given by

2 Z L L nπx bn = M0δ(x − ) sin dx L 0 2 L 2M = 0 sin (nπ/2) L  2M /L if n = 1, 5, 9, . . . ;  0 = −2M0/L if n = 3, 7, 11, . . . ;   0 if n = 2, 4, 6, . . . .

With bn determined, the solution for u(x, t) is given by

∞ 2M (2n + 1)πx 2 2 2 u(x, t) = 0 ∑ (−1)n sin e−(2n+1) π Dt/L . L n=0 L

When t L2/D, the leading-order term in the series is a good approximation and is given by

2M 2 2 u(x, t) ≈ 0 sin (πx/L)e−π Dt/L . L

The mass of the dye in the pipe is decreasing in time, diffusing into the reservoirs located at both ends. The total mass in the pipe at time t can be found from

Z L M(t) = u(x, t)dx, 0

and when t L2/D, we have 4M 2 2 M(t) = 0 e−π Dt/L . π

177 178 LECTURE 56. DIFFUSION EQUATION: EXAMPLE

Problems for Lecture 56

1. Determine the concentration of a dye in a pipe of length L at position x and time t, where the dye is initially concentrated uniformly across the center of the pipe with total mass M0, and the ends of the pipe are closed. Solutions to the Problems Practice quiz: The diffusion equation nπ 2 1. The solution of T0 + λDT = 0 with eigenvalues λ = results in the eigenfunctions n L nπDt a) T = cos n L nπDt b) T = sin n L

n2π2Dt c) T = exp n L2

n2π2Dt d) T = exp − n L2

∞ 2 2 2 2. If u(x, t) = a0/2 + ∑ an cos (nπx/L) exp (−n π Dt/L ) and u(x, 0) = f (x), the general formula n=1 for an is given by 2 Z L nπx a) an = f (x) cos dx L 0 L 2 Z L nπx b) an = f (x) sin dx L 0 L 2 Z L nπx π2D c) an = f (x) cos exp − dx L 0 L L2

2 Z L nπx π2D d) an = f (x) sin exp − dx L 0 L L2 3. Suppose that a pipe with open ends has an initial dye concentration centered at one-quarter of the pipe’s length. The long-time t >> L2/D solution for the concentration is given by √ 2M πx π2Dt a) u(x, t) = 0 sin exp − 2L L L2 √ 2M πx π2Dt b) u(x, t) = 0 sin exp − L L L2

2M πx π2Dt c) u(x, t) = 0 sin exp − L L L2 √ 2 2M πx π2Dt d) u(x, t) = 0 sin exp − L L L2 Solutions to the Practice quiz

179 180 LECTURE 56. DIFFUSION EQUATION: EXAMPLE Appendix

181

Appendix A

Complex numbers View this lecture on YouTube We deﬁne the imaginary unit i to be one of the two numbers (the other being −i) that satisﬁes z2 + 1 = √ 0. Formally, we write i = −1. A complex number z and its complex conjugate z¯ are written as

z = x + iy, z¯ = x − iy, where x and y are real numbers. We call x the real part of z and y the imaginary part, and write

x = Re z, y = Im z.

A linear combination of z and z¯ can be used to construct the real and imaginary parts of a number:

1 1 Re z = (z + z¯) , Im z = (z − z¯) . 2 2i

We can add, subtract and multiply complex numbers (using i2 = −1) to get new complex numbers. Division of two complex numbers is simpliﬁed by writing z/w = zw/ww. The exponential function of a complex number can be determined from a Taylor series. We have

(iθ)2 (iθ)3 (iθ)4 (iθ)5 eiθ = 1 + (iθ) + + + + ... 2! 3! 4! 5! θ2 θ4 θ3 θ5 = 1 − + − ... + i θ − + + ... = cos θ + i sin θ. 2! 4! 3! 5!

y z=x+iy

Im(z) r

x Re(z)

The complex plane.

The complex number x + iy = z can be represented in the complex plane with Re z as the x-axis and Im z as the y-axis (see the ﬁgure). Then with x = r cos θ and y = r sin θ, we have z = x + iy = r(cos θ + i sin θ) = reiθ.

183 184 APPENDIX A. COMPLEX NUMBERS Appendix B

Nondimensionalization

View this lecture on YouTube To nondimensionalize an equation, ﬁrst determine its base dimensions. These may be one or several of the fundamental dimensions such as time, length, mass and charge. The Buckingham π theorem states that after nondimensionalization, the remaining number of dimensionless parameters is the number of dimensional parameters minus the number of base dimensions. In practice, nondimensionalizing an equation can transform equations arising from seemingly different physical problems into an identical dimensionless equation, and can also signiﬁcantly reduce the number of parameters that must be explored in a numerical solution. To illustrate nondimensionalization, we consider the pendulum equation given by

mlθ¨ + clθ˙ + mg sin θ = F0 cos ωt,

where the four terms of the equation are called the inertial term, the frictional force, the restoring force, and the external force, respectively. The base dimensions are time (t), length (l) and mass (m). The angle θ expressed in radians is dimensionless, since it is deﬁned as the ratio of two lengths (arc

length divided by radius). The dimensional parameters are m, l, c, g, F0, and ω, for a total of six. The Buckingham π theorem says that after nondimensionalizing the equation, there should be no more than 6 − 3 = 3 dimensionless parameters. Each term in this equation must have the same dimension, and from the ﬁrst term we see that each term must have dimension ml/t2, which is the units of force. The frictional parameter denoted by c must have units m/t. The gravitational acceleration g has units of l/t2. If we divide the pendulum equation by ml, and deﬁne the natural frequency of the oscillator p without damping to be ω0 = g/l, then the pendulum equation becomes

c F θ¨ + θ˙ + ω2 sin θ = 0 cos ωt; m 0 ml

and the units of the various grouping of parameters are given by

c F [ ] = 1/t, [ω ] = 1/t, [ 0 ] = 1/t2, [ω] = 1/t. m 0 ml

At this stage there are four dimensional parameters and one base dimension, and nondimensionalization requires only the choice of a time scale. We see that there is some freedom in the choice of time scale, and we ﬁrst proceed without losing generality by deﬁning the dimensionless time by τ = t/t∗, where the time scale t∗ will be chosen later from our dimensional parameters. −1 2 2 −2 2 2 Using d/dt = (dτ/dt)d/dτ = t∗ d/dτ and d /dt = t∗ d /dτ , and multiplying the equation by

185 186 APPENDIX B. NONDIMENSIONALIZATION

2 t∗, the pendulum equation transforms to

2 2 d θ ct∗ dθ 2 2 F0t∗ + + ω t sin θ = cos (ωt∗τ). dτ2 m dτ 0 ∗ ml

A standard choice for t∗ arises from equating the coefficient of the inertial term to the coefficient of 2 2 the restoring force, that is, 1 = ω0t∗. This defines t∗ = 1/ω0. The resulting dimensionless equation becomes d2θ dθ + α + sin θ = γ cos βτ, dτ2 dτ where the three dimensionless parameters are given by

c ω F α = , β = , γ = 0 . mω0 ω0 mg

Another choice for nondimensionalization is to equate the coefficient of the frictional term and the 2 2 2 coefficient of the restoring force, that is, ct∗/m = ω0t∗. This defines t∗ = c/(mω0). The resulting 2 2 2 dimensionless equation after multiplication by m ω0/c becomes

d2θ dθ α + + sin θ = γ cos βτ, dτ2 dτ

where here m2g lcω F α = , β = , γ = 0 . lc2 mg mg This latter dimensionless equation is useful when studying very large damping effects, corresponding to the limit α → 0. The resulting governing equation then becomes ﬁrst-order in time. Appendix C

Matrices and determinants View this lecture on YouTube A two-by-two matrix A, with two rows and two columns, can be written as ! a a A = 11 12 . a21 a22

The first row has elements a11 and a12; the second row has elements a21 and a22. The first column has elements a11 and a21; the second column has elements a12 and a22. Matrices can be multiplied by scalars and added. This is done element-by-element and the straightforward definitions are ! ! ! ! ! a a ka ka a a b b a + b a + b k 11 12 = 11 12 , 11 12 + 11 12 = 11 11 12 12 . a21 a22 ka21 ka22 a21 a22 b21 b22 a21 + b21 a22 + b22

Matrices can also be multiplied. Matrix multiplication does not commute, and two matrices can be multiplied only if the number of columns of the matrix on the left equals the number of rows of the matrix on the right. One multiplies matrices by going across the rows of the ﬁrst matrix and down the columns of the second matrix. The two-by-two example is given by ! ! ! a a b b a b + a b a b + a b 11 12 11 12 = 11 11 12 21 11 12 12 22 . a21 a22 b21 b22 a21b11 + a22b21 a21b12 + a22b22

A system of linear algebraic equations can be written in matrix form. For instance, with aij and bi given numbers, and xi unknowns, the (two-by-two) system of equations given by

a11x1 + a12x2 = b1, a21x1 + a22x2 = b2,

can be written in matrix form as Ax = b, or explicitly as ! ! ! a a x b 11 12 1 = 1 . a21 a22 x2 b2

A unique solution to Ax = b exists only when det A 6= 0, and it is possible to write this unique solution as x = A−1b, where A−1 is the inverse matrix satisfying A−1A = AA−1 = I. Here, I is the identity matrix satisfying AI = IA = A. For the two-by-two matrix, the determinant is given by the product of the diagonal elements minus the product of the off-diagonal elements:

det A = a11a22 − a12a21.

187 188 APPENDIX C. MATRICES AND DETERMINANTS Appendix D

Eigenvalues and eigenvectors View this lecture on YouTube The eigenvalue problem for an n-by-n matrix A is given by

Ax = λx, where the scalar λ is called the eigenvalue and the n-by-1 column vector x is called the eigenvector. Using the identity matrix I, we can rewrite the eigenvalue equation as

(A − λI)x = 0.

When A is a two-by-two matrix, then ! ! 1 0 a − λ a I = and A − λI = 11 12 . 0 1 a21 a22 − λ

A solution other than x = 0 of the eigenvalue equation exists provided

det (A − λI) = 0.

This equation is called the characteristic equation of A, and is given by

2 λ − (a11 + a22)λ + (a11a22 − a12a21) = 0,

which is usually written as λ2 − TrA λ + detA = 0,

where TrA is the trace of the matrix A, equal to the sum of the diagonal elements. The eigenvalues can be real and distinct, complex conjugates, or repeated.

After determining an eigenvalue, say λ = λ1, the corresponding eigenvector v1 can be found by solving

(A − λ1I)v1 = 0,

or explicitly in the two-by-two case, ! ! ! a − λ a v 0 11 1 12 11 = . a21 a22 − λ1 v21 0

The second equation represented above is always a mutiple of the ﬁrst equation, and the eigenvector is unique only up to multiplication by a constant.

189 190 APPENDIX D. EIGENVALUES AND EIGENVECTORS Appendix E

Partial derivatives View this lecture on YouTube For a function f = f (x, y) of two variables, the partial derivative of f with respect to x is deﬁned as

∂ f f (x + h, y) − f (x, y) = lim , ∂x h→0 h

and similarly for the partial derivative of f with respect to y. To take the partial derivative of f with respect to x, say, take the derivative of f with respect to x holding y ﬁxed. As an example, consider

f (x, y) = 2x3y2 + y3.

We have ∂ f ∂ f = 6x2y2, = 4x3y + 3y2. ∂x ∂y Second derivatives are deﬁned as the derivatives of the ﬁrst derivatives, so we have

∂2 f ∂2 f = 12xy2, = 4x3 + 6y; ∂x2 ∂y2 and the mixed second partial derivatives are independent of the order in which the derivatives are taken: ∂2 f ∂2 f = 12x2y, = 12x2y. ∂x∂y ∂y∂x Partial derivatives are necessary for applying the chain rule. Consider d f = f (x + dx, y + dy) − f (x, y). We can write d f as

∂ f ∂ f d f = [ f (x + dx, y + dy) − f (x, y + dy)] + [ f (x, y + dy) − f (x, y)] = dx + dy. ∂x ∂y

If one has f = f (x(t), y(t)), say, then

d f ∂ f dx ∂ f dy = + . dt ∂x dt ∂y dt

And if one has f = f (x(r, θ), y(r, θ)), say, then

∂ f ∂ f ∂x ∂ f ∂y ∂ f ∂ f ∂x ∂ f ∂y = + , = + . ∂r ∂x ∂r ∂y ∂r ∂θ ∂x ∂θ ∂y ∂θ

191 192 APPENDIX E. PARTIAL DERIVATIVES Appendix F

Table of Laplace transforms

193 194 APPENDIX F. TABLE OF LAPLACE TRANSFORMS

f (t) = L−1{F(s)} F(s) = L{ f (t)}

1. eat f (t) F(s − a)

1 2. 1 s 1 3. eat s − a n! 4. tn sn+1 n! 5. tneat (s − a)n+1 b 6a. sin bt s2 + b2 b 6b. sinh bt s2 − b2 s 7a. cos bt s2 + b2 s 7b. cosh bt s2 − b2 b 8. eat sin bt (s − a)2 + b2 s − a 9. eat cos bt (s − a)2 + b2 2bs 10. t sin bt (s2 + b2)2 s2 − b2 11. t cos bt (s2 + b2)2 e−cs 12. u (t) c s −cs 13. uc(t) f (t − c) e F(s)

14. δ(t − c) e−cs

15. x˙(t) sX(s) − x(0)

16. x¨(t) s2X(s) − sx(0) − x˙(0)

Table of Laplace Transforms Appendix G

Problem and practice quiz solutions

195 196 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Classify differential equations

1. c, d, g. Third order, ordinary, nonlinear.

2. b, c , f. Second order, ordinary, nonlinear.

3. b, d , f. Second order, partial, nonlinear.

4. b, c , e. Second order, ordinary, linear.

5. b, d , e. Second order, partial, linear. 197

Solutions to the Problems for Lecture 2

1. The modiﬁed Euler method is

1 k = ∆x f (x , y ), k = ∆x f (x + ∆x, y + k ), y + = y + (k + k ). 1 n n 2 n n 1 n 1 n 2 1 2

The midpoint method is

1 1 k = ∆x f (x , y ), k = ∆x f (x + ∆x, y + k ), y + = y + k . 1 n n 2 n 2 n 2 1 n 1 n 2 198 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 3

1. 1 x2 − 4 a) dy = dx y x + 4 b) ey cos y dy = ex(1 + x) dx y + 1 x c) dy = dx y (x + 1) 1 d) dθ = −dt sin θ 199

Solutions to the Problems for Lecture 4

1. Separate the variables, integrate, and solve for the dependent variable. y x Z y dy Z x a = x dx y1/2 = x2 y1/2 − = x2 y = ( + x2)2 ) 1/2 4 ; 2 2 ; 1 ; 1 . 1 y 0 1 0 Z x Z t x dx 1 1 1 x(1 − x0) b) = dt. Use = + . [ln x − ln (1 − x)] = t; ln = t; x x(1 − x) 0 x(1 − x) x 1 − x x (1 − x) 0 x0 0 x(1 − x ) x 0 = t = 0 e ; x −t . x0(1 − x) x0 + (1 − x0)e 200 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Separable ﬁrst-order odes

dy 1. c. = x1/2y1/2, y(1) = 0. dx Z y dy Z x 2 (x3/2 − 1)2 = x1/2 dx; 2y1/2 = (x3/2 − 1); y = . 0 y1/2 1 3 9

dy 2. a. x = y2, y(1) = 1. dx Z y dy Z x dx 1 1 1 = ; −( − 1) = ln x; = 1 − ln x; y = . 1 y2 1 x y y 1 − ln x

dy 3. b. = −(sin x)y, y(π/2) = 1. dx Z y dy Z x = − sin x dx; ln y = cos x; y = ecos x. 1 y π/2 201

Solutions to the Problems for Lecture 5

1. dy 1 sin x a) + y = ; dx x x dy b) + y = x. dx

2. Let z = 1/x. Then x = 1/z and dx/dt = (dx/dz)(dz/dt) = −(1/z2)dz/dt. The differential equation becomes −(1/z2)dz/dt = (1/z)(1 − (1/z)), and after multiplying by −z2, dz/dt = −z + 1. In standard form, the linear ode is dz/dt + z = 1. 202 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 6

1. dy a) = x − y, y(0) = −1. dx

R x dx dy/dx + y = x; µ = e 0 = ex;

= −x(− + R x x ) = −x(− + x x − R x x ) = −x(− + x − x + ) = − y e 1 0 xe dx e 1 xe 0 0 e dx e 1 xe e 1 x 1. dy b) = 2x(1 − y), y(0) = 0. dx

R x 2x dx 2 dy/dx + 2xy = 2x; µ = e 0 = ex ;

2 −x2 R x x2 −x2 R x u −x2 x2 −x2 y = e 0 2xe dx = e 0 e du = e (e − 1) = 1 − e . 203

Solutions to the Practice quiz: Linear ﬁrst-order odes

1. d. (1 + x2)y0 + 2xy = 2x, y(0) = 0,

Z x 2 d h i x x (1 + x2)y = 2x, (1 + x2)y = 2x dx = x2, (1 + x2)y = x2, y = . dx 0 0 1 + x2

2. b. x2y0 + 2xy = 1, y(1) = 2,

Z x d h i x 1 + x x2y = 1, x2y = x2y − 2 = dx = x − 1, x2y = 1 + x, y = . dx 1 1 x2

3. d. y0 + λy = a, y(0) = 0,

Z x ae−λx a µ = eλx, y = e−λx aeλxdx = eλx − 1 = 1 − e−λx . 0 λ λ 204 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 7 k 1. Use S(t) = S ert + ert(1 − e−rt), with the unit of time in years, and money in US$. Here, t = 40, 0 r S0 = 0, r = 0.06, and S(t) =1,000,000. Solving for k using S0 = 0, we have

rS(t) 0.06 × 1,000,000 k = = = $5,986 per year. ert − 1 e0.06×40 − 1

The total amount saved is (40 years) × ($5,986 per year) ≈ $240,000, and the amount passively earned from the investment is $760,000, about three times as much that was saved.

k 2. Use S(t) = S ert + ert(1 − e−rt), with the unit of time in years, and money in US$. Here, S(t) is the 0 r amount owed the bank at time t, S0 is the total amount borrowed, r = 0.04 is the annual interest rate, and k = −1500 × 12 is the annual amount of repayment. We need to solve for S0 under the condition k that S(T) = 0 when T = 30 (loan repaid after 30 years). We have 0 = S erT + erT(1 − e−rT), or 0 r k 1500 × 12 S = − (1 − e−rT) = (1 − e−0.04×30) = $314, 463 ≈ $315, 000. 0 r 0.04 205

Solutions to the Problems for Lecture 8

1. Use mg v(t) = − (1 − e−kt/m), k where mg/k = 200 km/hr and m = 100 kg. With

g = (9.8 m/s2)(10−3 km/m)(60 s/min)2(60 min/hr)2 = 127, 008 km/hr2, we ﬁnd k = 63, 504 kg/hr. One-half of the terminal speed for free-fall (100 km/hr) is therefore attained when mg v(t) = − , 2k or (1 − e−kt/m) = 1/2, or t = m ln 2/k ≈ 4 sec. For 95% of the terminal speed (190 km/hr ), we have

(1 − e−kt/m) = 0.95,

and this equation is satisﬁed when t ≈ 17 sec. 206 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 9

−t/RC 1. The current i = dq/dt, with q = CVC. During charging, VC(t) = E 1 − e and during −t/RC −t/RC discharging VC(t) = Ee . With i = CdVC/dt, we have during charging, i = (E/R)e , and during discharging, i = −(E/R)e−t/RC. The currents strengths are the same during both processes but are in opposite directions. 207

Solutions to the Practice quiz: Applications

k 1. a. The relevant formula is S(T) = erT(1 − e−rT), where S(T) is the amount in the account after T r rS(T)e−rT years, r is the annual return, and k is the annual deposit rate. Solving for k, we have k = . 1 − e−rT Substituting r = 0.1, S(T) = 1,000,000, T = 40, we ﬁnd k =1,865.74. The total saved is 40k = 74,629.44 ≈ $75,000.

mg 2. c. Use v(t) = − (1 − e−kt/m), where mg/k = 200 km/hr, m = 100 kg and v(t) = −150 km/hr. k With g = 127,008 km/hr2, we ﬁnd k = 63,504 kg/hr. Three-quarter of the terminal speed for free-fall (150 km/hr) is therefore attained when (1 − e−kt/m) = 3/4, or t = m ln 4/k = 7.86 sec ≈ 8 sec.

V 3. c. Use V = E 1 − e−t/RC . Solve for t to ﬁnd t = −RC ln 1 − C . With R = 3000 Ω, C = 0.001 C E F, VC/E = 0.95, we ﬁnd t = 8.99 s ≈ 9 s. 208 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 10

1. θ˙ = u, u˙ = −u/q − sin θ + f cos ωt

2. k1 = ∆t f (tn, xn, yn), l1 = ∆tg(tn, xn, yn)

k2 = ∆t f (tn + ∆t, xn + k1, yn + l1), l2 = ∆tg(tn + ∆t, xn + k1, yn + l1) 1 1 x + = x + (k + k ), y + = y + (l + l ). n 1 n 2 1 2 n 1 n 2 1 2 209

Solutions to the Problems for Lecture 11

1. With x = xh(t) + xp(t), we compute

x¨ + p(t)x˙ + q(t)x = (x¨h + x¨p) + p x˙h + x˙ p + q xh + xp

= (x¨h + px˙h + qxh) + (x¨p + px˙ p + qxp) = 0 + g(t) = g(t), since xh and xp were assumed to be solutions of the homoogeneous and inhomogeneous equations, respectively. Therefore, their sum is a solution of the inhomogeneous equation. 210 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 12

1. With X1(t) = exp (αt) and X2(t) = exp (βt), we have

W = (exp (αt))(β exp (βt)) − (α exp (αt))(exp (βt)) = (β − α) exp ((α + β)t), which is nonzero provided α 6= β. 211

Solutions to the Problems for Lecture 13

a) r2 − 1 = 0, so that r = ±1;

b) r2 + 1 = 0, so that r = ±i;

c) r2 − 2r + 1 = 0, so that r = 1. 212 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Superposition, the Wronskian, and the characteristic equation

1. c. Functions that differ by a multiplying constant have a zero Wronskian. In particular, et−t0 = (e−t0 )et and sin (t − π) = − sin t. However, sin (t − π/2) = − cos t which has a nonzero Wronskian with sin t.

2. d. The principle of superposition applies so that we can multiply solutions by constants and add them to obtain a solution. The only proposed function that does not follow the principle of superposition is the product of two solutions.

3. c. An ode given by ax¨ + bx˙ + cx = 0 has complex-conjugate roots if b2 − 4ac < 0. In all four equations, b2 − 4ac = 1 − 4c and c = ±1, so only the equations with c = 1 have complex-conjugate roots. 213

Solutions to the Problems for Lecture 14

2 1. The characteristic equation is r + 4r + 3 = (r + 1)(r + 3) = 0, with roots r1 = −1 and r2 = −3. The −t −3t −t −3t general solution is x(t) = c1e + c2e . The derivative is x˙(t) = −c1e − 3c2e . Initial conditions are satisﬁed by solving c1 + c2 = 1 and −c1 − 3c2 = 0. Solution is c1 = 3/2 and c2 = −1/2. The ﬁnal 3 1 3 1 solution is x(t) = e−t − e−3t = e−t(1 − e−2t). 2 2 2 3

2 2. The characteristic equation is r − 1 = 0, with roots r1 = 1 and r2 = −1. The general solution t −t t −t is x(t) = c1e + c2e . The derivative is x˙(t) = c1e − c2e . Initial conditions are satisﬁed by solving c1 + c2 = x0 and c1 − c2 = u0. Solution is c1 = (x0 + u0)/2 and c2 = (x0 − u0)/2. The ﬁnal solution is

x + u x − u et + e−t et − e−t x(t) = 0 0 et + 0 0 e−t = x + u . 2 2 0 2 0 2

We can deﬁne et + e−t et − e−t cosh t = , sinh t = , 2 2 and write the solution as

x(t) = x0 cosh t + u0 sinh t.

The function cosh t is called the hyperbolic cosine of t, and is pronounced ‘cosh tee’, where the ‘o‘ sounds like the o in open. The function sinh t is called the hyperbolic sine of t and is pronounced ‘sinsh tee’, where the i sounds like the i in inch. 214 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 16

2 1. The characteristic equation is r − 2r + 5 = 0, with roots r1 = 1 + 2i and r2 = r1. The general solution is

x(t) = et (A cos 2t + B sin 2t) , x˙(t) = et (A cos 2t + B sin 2t) + 2et (−A sin 2t + B cos 2t) .

The initial conditions x(0) = 1 yields A = 1 and x˙(0) = 0 yields A + 2B = 0, or B = −1/2. The solution is 1 x(t) = et cos 2t − sin 2t . 2

2 2. The characteristic equation is r + 1 = 0, with roots r1 = i and r2 = −i. The general solution is x(t) = A cos t + B sin t. The derivative is x˙(t) = −A sin t + B cos t. Initial conditions are satisﬁed by

A = x0 and B = u0. The ﬁnal solution is x(t) = x0 cos t + u0 sin t. 215

Solutions to the Problems for Lecture 18

1. The characteristic equation is r2 − 2r + 1 = (r − 1)2 = 0, with repeated root r = 1. The general t t solution is x(t) = (c1 + c2t)e . The derivative is x˙(t) = (c1 + c2(1 + t))e . Initial conditions are satisﬁed t by solving c1 = 1 and c1 + c2 = 0. Solution is c1 = 1 and c2 = −1. Final solution is x(t) = (1 − t)e . 216 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Homogeneous equations

2 1. b. The characteristic equation is r − 3r + 2 = 0, with roots r1 = 2, r2 = 1. The general solution is 2t t 2t t x(t) = c1e + c2e , with derivative x˙(t) = 2c1e + c2e . Initial conditions are satisfied by solving x(0) = c1 + c2 = 1 and x˙(0) = 2c1 + c2 = 0 to find c1 = −1, c2 = 2. The final solution is x(t) = −e2t + 2et = −e2t(1 − 2e−t).

2 2. b. The characteristic equation is r − 2r + 2 = 0, with complex roots r± = 1 ± i. The general solution is x(t) = et (A cos t + B sin t), with derivative x˙(t) = et ((A + B) cos t + (B − A) sin t). Initial conditions are satisﬁed by A = 1 and A + B = 0, or B = −1. The ﬁnal solution is x(t) = et (cos t − sin t).

3. a. The characteristic equation is r2 + 2r + 1 = 0, with repeated root r = −1. The general solution is −t −t x(t) = e (c1 + c2t), with derivative x˙(t) = e (c2 − c1 − c2t). Initial conditions are satisﬁed by c1 = 0 −t and c2 − c1 = 1, or c2 = 1. The ﬁnal solution is x(t) = te . 217

Solutions to the Problems for Lecture 19

1. Substituting x(t) = xh(t) + xp1 (t) + xp2 (t) into the inhomogeneous ode, we obtain

d2 d x¨ + px˙ + qx = (x + x + x ) + p (x + x + x ) + q(x + x + x ) dt2 h p1 p2 dt h p1 p2 h p1 p2

= (x¨h + px˙h + qxh) + (x¨p1 + px˙ p1 + qxp1 ) + (x¨p2 + px˙ p2 + qxp2 )

= 0 + g1 + g2

= g1 + g2, so that the sum of the homogeneous and the two particular solutions solve the ode, and the two free constants in xh can be used to satisfy the two initial conditions. 218 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 20

1. The homogeneous equation has characteristic equation r2 + 5r + 6 = (r + 2)(r + 3) = 0, with roots −2t −3t r1 = −2 and r2 = −3. Therefore, xh = c1e + c2e . To ﬁnd the particular solution, we try the ansatz x = Ae−t. The resulting equation is A − 5A + 6A = 1, with solution A = 1/2. The general solution is therefore

1 1 x = c e−2t + c e−3t + e−t, x˙ = −2c e−2t − 3c e−3t − e−t. 1 2 2 1 2 2

Satisfying the inititial conditions results in the two equations c1 + c2 = −1/2 and −2c1 − 3c2 = 1/2, with solution c1 = −1 and c2 = 1/2. The ﬁnal solution is

1 1 1 x(t) = e−t − e−2t + e−3t = e−t 1 − 2e−t + e−2t . 2 2 2 219

Solutions to the Practice quiz: Solving inhomogeneous equations

1. a. For all three questions, we need to solve x¨ + 5x˙ + 6x = 2e−t. The homogeneous equation has 2 characteristic equation given by r + 5r + 6 = (r + 2)(r + 3) = 0, with roots r1 = −2 and r2 = −3. The −2t −3t general homogeneous solution is therefore xh = c1e + c2e . To ﬁnd a particular solution, we try −t xp = Ae . Substitution into the inhomogeneous ode yields A − 5A + 6A = 2, or A = 1. Therefore, the general solution to the inhomogeneous ode is the sum of the general homogeneous solution and the particular solution, or −2t −3t −t x(t) = c1e + c2e + e .

The three questions then entail solving for c1 and c2.

Here x(0) = 0 and x˙(0) = 0. We obtain the system of equations

c1 + c2 = −1

−2c1 − 3c2 = 1, with solution c1 = −2 and c2 = 1. Therefore,

x(t) = −2e−2t + e−3t + e−t = e−t(1 − 2e−t + e−2t).

2. a. Here x(0) = 1 and x˙(0) = 0. We obtain the system of equations

c1 + c2 = 0

−2c1 − 3c2 = 1, with solution c1 = 1 and c2 = −1. Therefore,

x(t) = e−2t − e−3t + e−t = e−t(1 + e−t − e−2t).

3. a. Here x(0) = 0 and x˙(0) = 1. We obtain the system of equations

c1 + c2 = −1

−2c1 − 3c2 = 2, with solution c1 = −1 and c2 = 0. Therefore,

x(t) = −e−2t + e−t = e−t(1 − e−t). 220 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 22

a) Try x(t) = A cos t + B sin t. Equating the coefﬁcients of cosine and sine, we obtain 5A + 3B = −2 and 3A − 5B = 0. The solution is A = −5/17 and B = −3/17.

b) Solve z¨ − 3z˙ − 4z = 2eit. Try z(t) = Ceit. Determine C = (−5 + 3i)/17 and ﬁnd

1 1 x = Re{z } = Re (−5 + 3i)(cos t + i sin t) = − (5 cos t + 3 sin t). p p 17 17 221

Solutions to the Problems for Lecture 23

1. Try x = At + B. Equating the coefﬁcients of t1 and t0, we obtain A = 1 and A + B = 0, or B = −1.

The solution is xp = t − 1. 222 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Particular solutions

2t 1. b. Try xp = Ae . Equating the coefﬁcients of the exponential function, we obtain 4A + 6A + 2A = 2, 1 with solution A = 1/6. The particular solution is given by x (t) = e2t. p 6

2it 2it 2. d. Solve the complex differential equation z¨ − z˙ − 2z = 2e , and take xp = Re{zp}. Try zp = Ce . −3 + i Equating the coefﬁcients of the exponential function, we obtain−4C − 2iC − 2C = 2, or C = . 10 1 1 We ﬁnd x (t) = Re{(−3 + i)(cos 2t + i sin 2t)} = − (3 cos 2t + sin 2t). p 10 10

3. c. Try xp = At + B to obtain −3A + 2At + 2B = t + 1. Equating coefﬁcients of powers of t, we ﬁnd 1 5 1 5 2A = 1 and −3A + 2B = 1, or A = , B = . The particular solution is given by x (t) = t + . 2 4 p 2 4 223

Solutions to the Problems for Lecture 24

1. The homogeneous equation has characteristic equation r2 + 3r + 2 = (r + 1)(r + 2) = 0, with roots −t −2t r1 = −1 and r2 = −2. Therefore, xh = c1e + c2e . To ﬁnd the particular solution, we try the ansatz x = Ate−2t. The resulting equation yields A = −1. The general solution is therefore

−t −2t −t −2t x = c1e + (c2 − t)e , x˙ = −c1e + (−1 − 2c2 + 2t)e .

Satisfying the initial conditions results in the two equations c1 + c2 = 0 and −c1 − 2c2 − 1 = 0, with solution c1 = 1 and c2 = −1. The ﬁnal solution is

x(t) = e−t − (1 + t)e−2t. 224 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 25

1. Write the RLC circuit equation as

L d2q dq 1 E + + q = 0 cos ωt. R dt2 dt RC R

To remove the coefficient of the restoring force in the final dimensionless equation, we define the dimensionless time τ to be τ = t/RC.

After multiplication by RC, the RLC circuit equation becomes

L d2q dq + + q = E C cos ωRCτ. R2C dτ2 dτ 0

To remove the coefﬁcient of the inhomogeneous term, we can deﬁne the dimensionless charge Q to be

q Q = , E0C and the dimensionless equation becomes

d2Q dQ α + + Q = cos βτ, dτ2 dτ

with L α = , β = ωRC. R2C 225

Solutions to the Problems for Lecture 26

1. Write the mass on a spring equation as

m d2x dx k F + + x = 0 cos ωt. c dt2 dt c c

To remove the coefficient of the restoring force in the final dimensionless equation, we define the dimensionless time τ to be τ = kt/c.

After multiplication by c/k, the mass on a spring equation becomes

mk d2x dx F ωc + + x = 0 cos τ. c2 dτ2 dτ k k

To remove the coefﬁcient of the inhomogeneous term, we can deﬁne the dimensionless position X to be kx X = , F0 and the dimensionless equation becomes

d2X dX α + + X = cos βτ, dτ2 dτ with mk ωc α = , β = . c2 k 226 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 28

1. The long-time solution corresponds to the particular solution. We consider the complex ode

z¨ + αz˙ + z = eiβt,

iβt 2 with xp = Re{zp}. With the ansatz zp = Ae , we have −β A + iαβA + A = 1, or

1 (1 − β2) − iαβ A = = . (1 − β2) + iαβ (1 − β2)2 + α2β2

To determine the amplitude of the resulting oscillation, we make use of the polar form of a complex number and write q (1 − β2) − iαβ = (1 − β2)2 + α2β2 eiφ, where tan φ = −αβ/(1 − β2). The long-time solution is given by

( p 2 2 2 2 ) (1 − β ) + α β i(βt+φ) xp = Re e (1 − β2)2 + α2β2 1 = p cos (βt + φ); (1 − β2)2 + α2β2

and the amplitude of the resulting oscillation is therefore given by

1 amplitude = p . (1 − β2)2 + α2β2 227

Solutions to the Practice quiz: Applications and resonance

1. d. We solve x¨ + x˙ = 1, with x(0) = 0 and x˙(0) = 0. The homogeneous equation has characteristic 2 equation given by r + r = r(r + 1) = 0, with roots r1 = 0 and r2 = −1. The general homogeneous −t solution is therefore xh = c1 + c2e . A particular solution is easily guessed to be xp = t. The general solution to the inhomogeneous ode is therefore

−t x(t) = c1 + c2e + t.

With x(0) = 0 and x˙(0) = 0, we obtain the system of equations

c1 + c2 = 0

c2 = 1, so that c1 = −1 and c2 = 1. Therefore,

x(t) = −1 + e−t + t = (t − 1) + e−t.

2. b. We solve x¨ − x = cosh t, with x(0) = 0 and x˙(0) = 0. The hyperbolic sine and cosine functions d d satisfy cosh t = sinh t and sinh t = cosh t. Furthermore, sinh (0) = 0 and cosh (0) = 1. The gen- dt dt eral homogeneous solution is xh = A cosh t + B sinh t. Since the inhomogeneous term is a solution of the homogeneous equation, we try as our particular solution xp = Ct cosh t + Dt sinh t. The derivatives are x˙ p = C cosh t + D sinh t + Ct sinh t + Dt cosh t and x¨p = 2C sinh t + 2D cosh t + Ct cosh t + Dt sinh t. Substituting into the ode, we get 2C sinh t + 2D cosh t = cosh t. Therefore C = 0 and D = 1/2. The general solution is therefore

1 x(t) = A cosh t + B sinh t + t sinh t, 2 1 1 x˙(t) = A sinh t + B cosh t + sinh t + t cosh t. 2 2

1 With x(0) = 0 and x˙(0) = 0, we obtain A = 0 and B = 0. Therefore, x(t) = t sinh t. 2

3. c. The fully dimensional RLC circuit equation, mass on a spring equation, and low amplitude pendulum equation are given by

d2q dq 1 L + R + q = E cos ωt, dt2 dt C 0 d2x dx m + c + kx = F cos ωt, dt2 dt 0 d2θ dθ ml + cl + mgθ = F cos ωt. dt2 dt 0

Comparing the three equations, it is observed that the resistance R in the RLC circuit equation plays a role analgous to the frictional coefﬁcent c in the mass on a spring equation, and the pendulum equation. 228 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 29

Z ∞ 1. F(s) = e−st sin bt dt. Integrate by parts letting u = sin bt and dv = e−stdt to obtain 0 b Z ∞ F(s) = e−st cos bt dt. Integrate by parts a second time letting u = b cos bt and dv = (1/s)e−stdt s 0 b b2 b to obtain F(s) = − F(s). Solving for F(s), we obtain F(s) = . s2 s2 s2 + b2 229

Solutions to the Problems for Lecture 30

1. The Laplace transform of the ode is given by

2 a s X − sx0 − u0 + b (sX − x0) + cX = G.

Solving for X = X(s), we obtain

G(s) + (as + b)x + au X(s) = 0 0 . as2 + bs + c 230 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 31

1. The Laplace transform of the ode is

1 s2X + 5sX + 6X = , s + 1

with solution 1 X(s) = . (s + 1)(s + 2)(s + 3) A partial fraction expansion of X = X(s) is written as

1 a b c = + + , (s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3 where the unknown coefficients can be found by using the cover-up method. Multiply both sides by (s + 1) and set s = −1 to find a = 1/2. Multiply both sides by s + 2 and set s = −2 to find b = −1. Multiply both sides by s + 3 and set s = −3 to find c = 1/2. Proceeding to take the inverse Laplace transform using the table in AppendixF, we obtain

1 1 x(t) = e−t − e−2t + e−3t. 2 2 231

Solutions to the Practice quiz: The Laplace transform method

1. b. From line 9 of the Table of Laplace Transforms, we have a = −1 and b = π. The Laplace s + 1 transform is then read off the table as X(s) = . (s + 1)2 + π2

2. c. Use lines 16, 15, and 3 of the Table of Laplace Transforms to take the Laplace transform of the 1 1 ode: s2X + sX − 6X = . Solve for X = X(s) to ﬁnd X(s) = . s + 1 (s + 1)(s − 2)(s + 3)

1 a b c 3. c. X(s) = = + + . Using the cover-up method, we ﬁnd (s + 1)(s + 2)(s + 3) s + 1 s + 2 s + 3 a = 1/2, b = −1, c = 1/2. Then using line 3 of the Table of Laplace Transforms, we obtain 1 1 x(t) = e−t − e−2t + e−3t. 2 2 232 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 32

1. The step-down function is given by ( 1, t < c; 1 − uc(t) = 0, t ≥ c.

The step-up, step-down function is given by

 0, t < a;  ua(t) − ub(t) = 1, a ≤ t < b;   0, t ≥ b.

2. The Laplace transform is

Z ∞ −st L{uc(t) f (t − c)} = e uc(t) f (t − c) dt 0 Z ∞ = e−st f (t − c) dt c Z ∞ 0 = e−s(t +c) f (t0) dt0 0 Z ∞ 0 = e−cs e−st f (t0) dt0 0 = e−csF(s), where we have changed variables to t0 = t − c.

a)  t, if t < 1; f (t) = 1, if t ≥ 1

= t − u1(t) (t − 1) .

b) The Laplace transform is given by F(s) = L{ f (t)} = L{t} − L{u1(t) (t − 1)}. The Laplace transforms are found in AppendixF. We have

1 L{t} = (using line 4 with n = 1) s2 e−s L{u (t) (t − 1)} = e−sL{t} = . (using line 13 with c = 1 and line 4 with n = 1) 1 s2

1 − e−s The result is F(s) = . s2 233

Solutions to the Problems for Lecture 33 Z ∞ 1 Z ∞ 1 1 Z ∞ 1. For a > 0, we have f (x)δ(ax) dx = f (x0/a)δ(x0) dx0 = f (0) = f (x)δ(x) dx. There- −∞ a −∞ a a −∞ 1 fore δ(ax) = δ(x). For a < 0, the limits of integration reverse upon the substitution, and one ﬁnds a 1 1 δ(ax) = − δ(x). The general result is δ(ax) = δ(x). a |a|

Z x 2. The integral given by δ(x0 − c) dx0 is zero if x < c and one if x > c. This is the deﬁnition of −∞ uc(x).

Z x 0 0 3. The derivative of both sides of uc(x) = δ(x − c) dx with respect to x, using the fundamen- −∞ d tal theorem of calculus, results in u (x) = δ(x − c). This is only a symbolic expression because dx c technically you can not take the derivative of a discontinuous function. 234 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 34

1. We compute x(1) when the Heaviside function is either zero or one. In both cases we ﬁnd

1 1 1 x(1) = − + 2 e 2e2

so that x = x(t) is continuous at t = 1.

2. The Laplace transform is 1 s2X(s) + X(s) = 1 − e−2πs , s with solution for X = X(s) given by 1 − e−2πs X(s) = . s(s2 + 1) Deﬁning 1 F(s) = , s(s2 + 1) and using the table in AppendixF, the inverse Laplace transform of X(s) is

x(t) = f (t) − u2π(t) f (t − 2π), where f (t) is the inverse Laplace transform of F(s). A partial fraction expansion yields

1 1 s F(s) = = − , s(s2 + 1) s s2 + 1

and the inverse Laplace transform from the table in AppendixF is f (t) = 1 − cos t. We ﬁnd

x(t) = 1 − cos t − u2π(t) (1 − cos (t − 2π)) ,

or more clearly,  1 − cos t, if t < 2π; x(t) = 0, if t ≥ 2π. 235

Solutions to the Problems for Lecture 35

1. Taking the Laplace transform of the ode using the table in AppendixF, we have

s2X + X = 1 − e−2πs, with solution for X = X(s) given by 1 − e−2πs X(s) = . (s2 + 1) Deﬁning 1 F(s) = , s2 + 1 the table in AppendixF yields f (t) = sin t, and the inverse Laplace transform of X(s) is

x(t) = f (t) − u2π(t) f (t − 2π) = sin t − u2π(t) sin(t − 2π), or more clearly,  sin t, if t < 2π; x(t) = . 0, if t ≥ 2π. 236 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Discontinuous and impulsive inhomogeneous terms

1. a. Use the negative of a step-up, step-down function between t = 1 and 2, and a step-up, step-down

function between t = 3 and 4. We have x(t) = −(u1(t) − u2(t)) + (u3(t) − u4(t)) = −u1(t) + u2(t) + u3(t) − u4(t).

1 e−2πs 2. d. Laplace transform the ode to obtain s2X − s + X = − . Solve for X = X(s) to obtain s s s (1 − e−2πs) 1 1 s 1 1 s X(s) = + . Write = − , so that X(s) = − e−2πs − . s2 + 1 s(s2 + 1) s(s2 + 1) s s2 + 1 s s s2 + 1  1, if t < 2π; Inverse Laplace transform to ﬁnd x(t) = 1 − u2π(t) (1 − cos (t − 2π)) = cos t, if t ≥ 2π.

3. c. Laplace transform the ode to obtain s2X − s + X = 1 − e−2πs. Solve for X = X(s) to obtain s 1 e−2πs X(s) = + − . Inverse Laplace transform to ﬁnd s2 + 1 s2 + 1 s2 + 1  cos t + sin t, if t < 2π; x(t) = cos t + sin t − u2π(t) sin (t − 2π) = cos t, if t ≥ 2π. 237

Solutions to the Problems for Lecture 36

1. Try ∞ n y(x) = ∑ anx , n=0 and obtain ∞ ∞ n−2 n ∑ n(n − 1)anx − ∑ anx = 0. n=2 n=0 In the ﬁrst sum, shift the summation index downward by two, and then combine sums to obtain

∞ n ∑ (n + 2)(n + 1)an+2 − an x = 0. n=0

We therefore obtain the recursion relation

an a + = , n = 0, 1, 2, . . . . n 2 (n + 2)(n + 1)

Even and odd coefﬁcients decouple, and we obtain two independent sequences starting with ﬁrst term

a0 or a1. Developing these sequences, we have for the ﬁrst sequence,

1 1 1 a , a = a , a = a = a , 0 2 2 0 4 4 · 3 2 4! 0

and so on; and for the second sequence,

1 1 1 a , a = a , a = a = a , 1 3 3 · 2 1 5 5 · 4 3 5! 1

and so on. Using the principle of superposition, the general solution is therefore

x2 x4 x3 x5 y(x) = a 1 + + + ... + a x + + + ... 0 2! 4! 1 3! 5!

= a0 cosh x + a1 sinh x,

where ex + e−x ex − e−x cosh x = , sinh x = . 2 2 238 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 37

∞ n 1. Substituting y = ∑n=0 anx into the differential equation, we have

∞ ∞ ∞ 00 0 n−2 n n y + xy − y = ∑ n(n − 1)anx + ∑ nanx − ∑ anx n=2 n=1 n=0 ∞ n = ∑ (n + 2)(n + 1)an+2 + (n − 1)an x = 0. n=0

The recursion relation is (n − 1) a + = − a . n 2 (n + 2)(n + 1) n

Starting with a0, we ﬁnd a2 = a0/2, a4 = −a0/24, and a6 = a0/240. Starting with a1, we ﬁnd a3 = 0, 6 a5 = 0, etc. The general solution to order x is given by

x2 x4 x6 y(x) = a 1 + − + − ... + a x. 0 2 24 240 1 239

Solutions to the Problems for Lecture 38

1. The general solution of the Airy’s equation is given by

y(x) = a0y0(x) + a1y1(x).

0 0 Applying the initial conditions and the known values y0(0) = 1, y0(0) = 0, y1(0) = 0, and y1(0) = 1, we have 0 y(0) = 1 = a0, y (0) = 1 = a1.

Therefore, the solution is

y(x) = y0(x) + y1(x). 240 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Series solutions

1. c. Using the deﬁnitions of the hyperbolic functions, we have (ex + e−x)2 (ex − e−x)2 cosh2 t − sinh2 t = − 4 4 1 1 1 1 = e2x + e−2x + 2 − e2x + e−2x − 2 = + = 1. 4 4 2 2

∞ n 2. d. Substituting y = ∑ anx into the differential equation, we have n=0 ∞ ∞ ∞ ∞ 00 2 n−2 n+2 n+2 n+2 y + x y = ∑ n(n − 1)anx + ∑ anx = ∑ (n + 4)(n + 3)an+4x + ∑ anx = 0. n=2 n=0 n=−2 n=0 ∞ n+2 Rewrite as 2a2 + 6a3x + ∑ [(n + 4)(n + 3)an+4 + an] x = 0. Setting coefficients of powers of x n=0 an equal to zero results in a = a = 0, a + = − . Starting with a , we find a = −a /12. 2 3 n 4 (n + 4)(n + 3) 0 4 0 Starting with a1, we find a5 = −a1/20. Since a2 = a3 = a6 = a7 = 0, the general solution with terms x4 x5 up to x5 is y(x) = a 1 − + ... + a x − + ... . 0 12 1 20

∞ n 3. d. Substituting y = ∑ anx into the differential equation, we have n=0 ∞ ∞ ∞ ∞ 00 0 n−2 n n n y − xy + y = ∑ n(n − 1)anx − ∑ nanx + ∑ anx = ∑ (n + 2)(n + 1)an+2 + (1 − n)an x = 0. n=2 n=1 n=0 n=0 (n − 1) The recursion relation is a + = a . Starting with a , we ﬁnd a = −a /2, a = −a /24. n 2 (n + 2)(n + 1) n 0 2 0 4 0 4 Starting with a1, we ﬁnd a3 = 0, a5 = 0, etc. The general solution up to terms proportional to x is x2 x4 given by y(x) = a 1 − − − ... + a x. 0 2 24 1 241

Solutions to the Problems for Lecture 39

1. The odes in matrix form are given by ! ! ! d x a c x 1 = 1 , dt x2 c b x2 and the characteristic equation of the matrix is given by

λ2 − (a + b)λ + ab − c2 = 0.

The roots of the characteristic equation are

p p a + b ± (a + b)2 − 4ab + 4c2 a + b ± (a − b)2 + 4c2 λ± = = . 2 2

Since the discriminant is non-negative, both roots are real (and distinct provided a 6= b and c 6= 0). 242 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 40 ! 0 −1 1. With A = , the characteristic equation is λ2 + λ − 2 = (λ − 1)(λ + 2) = 0, with roots −2 −1 ! ! 1 1 λ1 = 1 and λ2 = −2. The corresponding eigenvectors are v1 = and v2 = . The general −1 2 ! ! t 1 −2t 1 t −2t t −2t solution is x = c1e + c2e , or x1 = c1e + c2e and x2 = −c1e + 2c2e . −1 2 243

Solutions to the Problems for Lecture 41 ! 1 −2 √ 1. With A = , the characteristic equation is λ2 − 2λ + 3 = 0, with root λ = 1 + i 2 1 1 ! 2i and its complex conjugate. The corresponding eigenvectors are v = √ and its complex conju- 2 n λto gate. The general solution is constructed from the linearly independent solutions X1 = Re ve and n λto X2 = Im ve , and is

√ ! √ !! −2 sin ( 2 t) 2 cos ( 2 t) x = et A √ √ + B √ √ , 2 cos ( 2 t) 2 sin ( 2 t) or √ √ √ √ √ √ t t x1 = e −2A sin ( 2 t) + 2B cos ( 2 t) , x2 = e 2A cos ( 2 t) + 2B sin ( 2 t) . 244 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Systems of differential equations ! ! ! d x a b x 1. c. The system of equations in matrix form is 1 = 1 . The characteristic equa- dt x2 c d x2 tion of the matrix is given by λ2 − (a + d)λ + (ad − bc) = 0. The second-order ode with the same characteristic equation is given by x¨ − (a + d)x˙ + (ad − bc)x = 0.

! ! ! d x 1 2 x 2. b. The system of equations in matrix form is 1 = 1 . Try x = veλt to obtain the dt x2 2 1 x2 eigenvalue problem Av = λv. The characteristic equation of A from det (A − λI) = 0 is λ2 − 2λ − 3 =

(λ − 3)(λ + 1) = 0, with roots λ1 = 3 and λ2 = −1. The eigenvectors are found by solving ! 1 (A − λiI)vi = 0, and we ﬁnd that λ1 = 3 has eigenvector v1 = and λ2 = −1 has eigenvector 1 ! ! ! 1 1 3t 1 −t v2 = . The general solution is x = c1 e + c2 e , and the components are x1 = −1 1 −1 3t −t 3t −t c1e + c2e and x2 = c1e − c2e .

! ! ! d x −2 1 x 3. d. The system of equations in matrix form is 1 = 1 . Try x = veλt to ob- dt x2 −1 −2 x2 tain the eigenvalue problem Av = λv. The characteristic equation of A from det (A − λI) = 0 is λ2 + 4λ + 5 = 0, with roots λ = −2 + i and its complex conjugate. The eigenvector associated ! 1 with λ is found by solving (A − λI)v = 0, and we ﬁnd that v1 = . The general solution is i ( ! ) ( ! )! ! !! 1 1 cos t sin t x = e−2t A Re eit + B Im eit = e−2t A + B , and the compo- i i − sin t cos t −2t −2t nents are x1 = e (A cos t + B sin t) and x2 = e (−A sin t + B cos t). 245

Solutions to the Problems for Lecture 42

1. The nature of the ﬁxed points are determined by the eigenvalues of the relevant matrices. We ﬁrst write the odes as matrix equations, and then compute the eigenvalues.

a) √ ! −3 2 x˙ = √ x. 2 −2

The eigenvalues of the matrix are found from

0 = det (A − λI) = λ2 + 5λ + 4 = (λ + 4)(λ + 1).

The eigenvalues are λ1 = −4 and λ2 = −1. Since the eigenvalues are both negative, the ﬁxed point is a stable node.

b) ! 1 1 x˙ = x. 4 1

The eigenvalues of the matrix are found from

0 = det (A − λI) = λ2 − 2λ − 3 = (λ + 1)(λ − 3).

The eigenvalues are λ1 = −1 and λ2 = 3. Since the eigenvalues are of opposite sign, the ﬁxed point is a saddle point and is unstable.

c) ! −1/2 1 x˙ = x. −1 −1/2

The eigenvalues of the matrix are found from

5 0 = det (A − λI) = λ2 + λ + . 4

1 The eigenvalues are complex and given by λ = − ± i. Since the real part of these complex 2 eigenvalues is negative, the ﬁxed point is a stable spiral point. 246 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 43 ! 2 1 1. With A = , the characteristic equation is λ2 − 4λ + 3 = (λ − 3)(λ − 1) = 0, with roots 1 2 ! ! 1 1 λ1 = 3 and λ2 = 1. The corresponding eigenvectors are v1 = and v2 = . The phase 1 −1 portrait can be sketched by drawing the lines x2 = x1 corresponding to the first eigenvector direction, and x2 = −x1 corresponding to the second eigenvector. Because both eigenvalues are positive, the motion is away from the origin and the fixed point is unstable. Also, λ1 > λ2, so the motion is faster along the first eigenvector with x2 = x1 and the curves bend in this direction. A computer-generated phase portrait is shown below.

1 2

x 0

-1

-2 -2 -1 0 1 2 x 1 247

Solutions to the Problems for Lecture 44 ! −1 3 1. With A = , the characteristic equation is λ2 − 3λ − 10 = (λ + 2)(λ − 5) = 0, with roots 2 4 ! ! 1 1 λ1 = −2 and λ2 = 5. The corresponding eigenvectors are v1 = and v2 = . The phase −1/3 2 portrait can be sketched by drawing the lines x2 = −x1/3 corresponding to the first eigenvector, and x2 = 2x1 corresponding to the second eigenvector. Because the eigenvalues have opposite sign, the fixed point is a saddle point. Also, λ1 < 0 < λ2 so the motion along the first eigenvector moves towards the origin, and the motion along the second eigenvector moves away from the origin. A computer-generated phase portrait is shown below.

1 2

x 0

-1

-2 -2 -1 0 1 2 x 1 248 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 45 ! 1 1 1. With A = , the characteristic equation is λ2 − 2λ + 2 = 0, with roots λ = 1 + i and −1 1 ! 1 λ¯ . The corresponding eigenvectors are v = and ¯v. The phase portrait is an unstable spiral i point. To determine the handedness, we compute the derivatives at the point (x1, x2) = (0, 1) and ﬁnd (x˙1, x˙2) = (1, 1). The stability and direction of motion indicates a clockwise spiral moving outwards. A computer-generated phase portrait is shown below.

1 2

x 0

-1

-2 -2 -1 0 1 2 x 1 249

Solutions to the Practice quiz: Phase portraits

1. a. See Problems for Lecture 43: Stable and unstable nodes.

2. b. See Problems for Lecture 44: Saddle points.

3. c. See Problems for Lecture 45: Spiral points. 250 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 46

1. ! ! ! d2 x −2k k x 1 = 1 m 2 . dt x2 k −k x2 251

Solutions to the Problems for Lecture 48

1. The governing equation may be written as ! ! ! m d2 x −2 1 x 1 = 1 2 . k dt x2 1 −1 x2

= rt = = 2 With ansatz x ve , the matrix equation becomes Av λv,√ with λ mr /k.The eigenvalues√ of A are 2 found from det(A − λI) = λ + 3λ + 1 = 0, or λ1 = −(3 − 5)/2 and λ2 = −(3 + 5)/2. The cor- √ T √ T responding eigenvectors are computed to be v1 = 1 ( 5 + 1)/2 and v2 = 1 −( 5 − 1)/2 . The approximate values of the angular frequencies and the eigenvectors are

r ! k 1 ω1 = 0.62 , v1 = ; m 1.62 r ! k 1 ω2 = 1.62 , v2 = . m −0.62 252 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Normal modes

1. c. Newton’s law and Hooke’s law results in mx¨ = −2kx − k(x − x ) and mx¨ = −k(x − x ). In ! 1 ! 1 ! 1 2 2 2 1 d2 x −3k k x 1 = 1 matrix form, this is written as m 2 . dt x2 k −k x2

! ! ! m d2 x −3 1 x 1 = 1 ( ) = rt 2. a. The governing equations are given by 2 . Try x t ve , and ob- k dt x2 1 −1 x2 2 mr 2 tain Av = λv, with λ = . The characteristic equation of A is λ + 4λ + 2 = 0, with solutions λ± = k q √ q √ r (2− 2)k k |λ±|k  m ≈ 0.77 m , −2 ± 2. The angular frequencies of the normal modes are given by = q √ q m  (2+ 2)k k m ≈ 1.85 m .

! −3 1 √ 3. c. We ﬁnd the eigenvectors of the matrix A = with eigenvalues λ1 = −2 + 2 and 1 −1 √ √ ! ! −1 − 2 1 v11 λ2 = −2 − 2. We solve (A − λiI)vi = 0 for vi. For λ1, we have √ = 0, 1 1 − 2 v21 √ ! ! √ −1 + 2 1 v12 or v21 = (1 + 2)v11 ≈ 2.41v11. For λ2, we have √ = 0, or v22 = (1 − 1 1 + 2 v22 ! ! √ 1 1 2)v12 ≈ −0.41v12. Therefore, v1 = and v2 = . 2.41 −0.41 253

Solutions to the Problems for Lecture 49

a) Computing f (x + 2L), we have

∞ a0 nπ(x + 2L) nπ(x + 2L) f (x + 2L) = + ∑ an cos + bn sin 2 n=1 L L ∞ a0 nπx nπx = + ∑ an cos + 2nπ + bn sin + 2nπ 2 n=1 L L ∞ a0 nπx nπx = + ∑ an cos + bn sin 2 n=1 L L = f (x).

b) Computing the average value of f (x), we have

1 Z L h f (x)i = f (x)dx 2L −L 1 Z L a ∞ a Z L nπx b Z L nπx = 0 dx + ∑ n cos dx + n sin dx 2L −L 2 n=1 2L −L L 2L −L L a = 0 . 2

Therefore, a0 = 2 h f (x)i. 254 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 50

1. If f (x) is an odd function, then its Fourier sine series is given by

∞ nπx f (x) = ∑ bn sin , n=1 L

∞ and f (0) = ∑ bn sin 0 = 0. If f (x) is an even function, then its Fourier cosine series is given by n=1

∞ a0 nπx f (x) = + ∑ an cos , 2 n=1 L and ∞ 0 π nπx f (x) = − ∑ nan sin . L n=1 L ∞ 0 π Therefore, f (0) = − ∑ nan sin 0 = 0. L n=1 255

Solutions to the Problems for Lecture 51

1. The function f (x) is odd and L = π. The coefﬁcients of the Fourier sine series are given by

2 Z π bn = sin nx dx π 0 2 = − cos nx|π nπ 0 2 = (1 − cos nπ) nπ  4 1, if n odd; = × nπ 0, if n even.

The Fourier sine series is therefore

4 sin 3x sin 5x f (x) = sin x + + + ... . π 3 5 256 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Fourier series

∞ ∞ a0 a0 1. b. f (0) = + ∑ (an cos 0 + bn sin 0) = + ∑ an. 2 n=1 2 n=1

2. c. The square wave in the ﬁgure is an odd function of period 2π, with f (x) = 1 for 0 < x < π. The ∞ Fourier sine series is given by f (x) = ∑ bn sin nx with n=1  Z π Z π π 4 2 2 2 cos nx  πn , n odd; bn = f (x) sin nx dx = sin nx dx = − = π 0 π 0 πn 0 0, n even. 4 sin 3x sin 5x Therefore, f (x) = sin x + + + ... . π 3 5

3. b. This is the function f (x) of Question 2 where f (x) is shifted to the left by π/2. If we denote this shifted function by g(x), then g(x) = f (x + π/2). We have

4 sin (3 (x + π/2)) sin (5 (x + π/2)) g(x) = sin (x + π/2) + + + ... . With sin (x + π/2) = cos x, π 3 5 4 cos 3x cos 5x sin (3 (x + π/2)) = − cos 3x, etc., we have g(x) = cos x − + − ... . π 3 5 257

Solutions to the Problems for Lecture 52

1. We consider ∂u ∂2u = D , ∂t ∂x2 and deﬁne the dimensionless time as τ = tD/L2 and the dimensionless position as s = x/L. Changing variables, the nondimensional diffusion equation becomes

∂u ∂2u = , ∂τ ∂s2 which is an equation without any parameters. 258 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 53

1. We have 0 0 ux(0, t) = X (0)T(t) = 0, ux(L, t) = X (L)T(t) = 0.

Since these boundary conditions are valid for all t, we must have X0(0) = X0(L) = 0. These are called homogeneous Neumann boundary conditions. 259

Solutions to the Problems for Lecture 54

1. If λ = 0, the general solution of X00 = 0 is given by

X(x) = A + Bx, and X(0) = 0 yields A = 0, and X(L) = 0 yields B = 0. If λ < 0, we write λ = −µ2, and determine the general solution of X00 − µ2X = 0

to be X(x) = Aeµx + Be−µx.

The boundary condition at x = 0 results in A + B = 0, and the boundary condition at X = L results in AeµL + Be−µL = 0. Combining these two equations yields A eµL − e−µL = 0.

The only solution for nonzero µ is given by A = 0, and since B = −A, there are no nontrivial solutions.

2. Nontrivial solutions only exist for λ ≥ 0. If λ = 0, the general solution of X00 = 0 is given by

X(x) = A + Bx,

and the boundary conditions X0(0) = X0(L) = 0 require B = 0. We can write the eigenfunction 2 associated with λ = 0 as X0(x) = 1. If λ > 0, we let λ = µ , and the general solution of the ode and its derivative is given by

X(x) = A cos µx + B sin µx, X0(x) = −µA sin µx + µB cos µx.

The boundary condition X0(0) = 0 requires B = 0 and the boundary condition X0(L) = 0 results in

sin µL = 0, with solutions given by µn = nπ/L, where n is a nonzero integer. The eigenvalues are 2 therefore λn = (nπ/L) for n = 1, 2, 3, . . . with corresponding eigenfunctions Xn(x) = cos (nπx/L). 260 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Practice quiz: Separable partial differential equations

−1 −2 1. d. The units on both sides of an equation must agree. We have [ut] = [D][uxx], or ut = [D]ul , where u denotes concentration, t denotes time, and l denotes length. Therefore, [D] = l2t−1.

2 00 2 00 2. d. To solve utt = c uxx, we substitute u(x, t) = X(x)T(t) to obtain XT = c X T. Separating, we obtain X00/X = T00/c2T = −λ, where λ is the separation constant. The two differential equations can be written as X00 + λX = 0 and T00 + λc2T = 0.

3. c. The differential equation X00 + λX = 0 with boundary conditions X(0) = 0 and X0(L) = 0 has nontrivial solutions only when λ > 0. We let λ = µ2 > 0, and solve X00 + µ2X = 0 to ﬁnd X(x) = A cos µx + B sin µx. Applying the boundary conditions, we have X(0) = 0 = A and X0(L) = 0 = (2n − 1)π µB cos µL. Since B 6= 0, we must have cos µL = 0. The allowed values of µ are given by µ = n 2L (2n − 1)π 2 for n = 1, 2, 3, . . . . The eigenvalues and eigenfunctions are therefore given by λ = and n 2L (2n − 1)πx X = sin , for n = 1, 2, 3, . . . . n 2L 261

Solutions to the Problems for Lecture 55

2 1. With the eigenvalues λn = (nπ/L) , for n = 0, 1, 2, 3, . . . , the differential equation for T = T(t) becomes T0 + n2π2D/L2 T = 0,

which has solution proportional to −n2π2Dt/L2 Tn = e .

Therefore, with ansatz u(x, t) = X(x)T(t) and eigenvalues λn, we conclude that the functions

−n2π2Dt/L2 un(x, t) = cos (nπx/L)e

satisfy the diffusion equation and the spatial boundary conditions for every nonnegative integer n. The principle of linear superposition for homogeneous linear differential equations then states that the general solution to the diffusion equation with the spatial boundary conditions is given by

∞ ∞ a0 a0 −n2π2Dt/L2 u(x, t) = + ∑ anun(x, t) = + ∑ an cos (nπx/L)e , 2 n=1 2 n=1

where following the usual convention we have separated out the n = 0 solution. In the final step, we assume that u(x, 0) = f (x), where f (x) is some specific function defined on 0 ≤ x ≤ L. At t = 0, we have ∞ a0 f (x) = + ∑ an cos (nπx/L). 2 n=1 We immediately recognize this equation as a Fourier cosine series for an even function f (x) with period 2L. A Fourier cosine series results because of the boundary condition f 0(0) = 0. From our previous solution for the coefficients of a Fourier cosine series, we determine that

2 Z L nπx an = f (x) cos dx. L 0 L 262 APPENDIX G. PROBLEM AND PRACTICE QUIZ SOLUTIONS

Solutions to the Problems for Lecture 56

1. We solve the diffusion equation with homogeneous Neumann boundary conditions, and model the

initial concentration of the dye by a delta-function centered at x = L/2, that is, u(x, 0) = M0δ(x − L/2). The Fourier cosine series coefﬁcients are therefore given by

2 Z L L nπx an = M0δ(x − ) cos dx L 0 2 L 2M = 0 cos (nπ/2) L  2M /L if n = 0, 4, 8, . . . ;  0 = −2M0/L if n = 2, 6, 10, . . . ;   0 if n = 1, 3, 5, . . . .

The coefficients an are nonzero only for even values of n, so we can redefine n and write the solution for u(x, t) as M 2M ∞ 2nπx 4n2π2Dt ( ) = 0 + 0 (− )n − u x, t ∑ 1 cos exp 2 . L L n=1 L L As one would expect for a pipe with closed ends, after a sufficiently long time, the mass of the dye becomes uniformly distributed across the length of the pipe. 263

Solutions to the Practice quiz: The diffusion equation

1. d. The solution of T0 + λDT for T = T(t) up to a multiplicative constant is T = exp (−λDt). n2π2Dt Replacing λ by the eigenvalue λ = (nπ/L)2, we obtain the eigenfunction T = exp − . n n L2

∞ 2 2 2 2. a. We have u(x, t) = a0/2 + ∑ an cos (nπx/L) exp (−n π Dt/L ) and u(x, 0) = f (x). Therefore, n=1 ∞ f (x) = a0/2 + ∑ an cos (nπx/L). This is a Fourier cosine series for f (x) and the coefﬁcients are given n=1 2 Z L nπx by an = f (x) cos dx . L 0 L

3. b. The solution for the concentration in a pipe with open ends is given by ∞ nπx n2π2Dt 2 Z L nπx ( ) = − ( ) = ( ) = ( ) u x, t ∑ bn sin exp 2 . With u x, 0 f x , we have bn f x sin dx. n=1 L L L 0 L Z L L 2 L nπx 2M0 nπ Here, f (x) = M0δ x − . We have bn = M0δ x − sin dx = sin . The 4 L 0 4 L √L 4 2M0 2M0 leading-order term corresponds to n = 1 where we have b1 = sin (π/4) = . Therefore √ L L 2M πx π2Dt u(x, t) = 0 sin exp − . L L L2